diff --git a/记忆题.md b/记忆题.md new file mode 100644 index 0000000..7ab5899 --- /dev/null +++ b/记忆题.md @@ -0,0 +1,2222 @@ +```python +#R 75 Sort Colors/颜色分类 (Medium) +# p0：指向 0 应该放置的位置；p0左边全是0, p0本身并不包括0 +# p2：指向 2 应该放置的位置；p2右边全是0, p2本身并不包括2 +# i：当前遍历位置 +class Solution: + def sortColors(self, nums: List[int]) -> None: + p0, i , p2 = 0,0,len(nums) - 1 + while i <= p2: + if nums[i] == 0: + nums[i], nums[p0] = nums[p0], nums[i] + p0 += 1 + i += 1 + elif nums[i] == 2: + nums[i], nums[p2] = nums[p2], nums[i] + p2 -= 1 + # 注意：i 不增加，因为换过来的 nums[i] 还要检查 + else: + i += 1 +``` + +```python +#R 259 3Sum Smaller $/较小的三数之和 (Medium) +# satisfy the condition nums[i] + nums[j] + nums[k] < target +class Solution: + def threeSumSmaller(self, nums, target): + if len(nums) < 3: + return 0 + res = 0 + nums.sort() + n = len(nums) + for i in range(n - 2): + left, right = i + 1, n - 1 + while left < right: + if nums[i] + nums[left] + nums[right] < target: + res += right - left # 所有 (nums[i], nums[left], nums[left+1]...nums[right]) 都满足条件 + left += 1 + else: + right -= 1 + return res +``` + +```python +#H 1163 Last Substring in Lexicographical Order/按字典序排在最后的子串 (Hard) +# 找出它的所有子串并按字典序排列，返回排在最后的那个子串。 +# i指向的是当前找到字典序最大的字符，j指向的是当前要进行比较的字符。使用一个位移指针k，来比较i和j构成的子串[i,..,i + k]和[j,...,j + k]的顺序。i始终指向当前找到字典序最大的字符 +class Solution: + def lastSubstring(self, s: str) -> str: + n = len(s) + i = 0 + j = 1 + k = 0 + while j + k < n: + if s[i + k] == s[j + k]: + k += 1 + elif s[i + k] < s[j + k]: + i += k + 1 + k = 0 + if i >= j: + j = i + 1 + else: + j += k + 1 + k = 0 + return s[i:] +``` + + +# 滑动窗口 + +```python +#R 1358 Number of Substrings Containing All Three Characters/包含所有三种字符的子字符串数目 (Medium) +# 请你返回 a，b 和 c 都 至少 出现过一次的子字符串数目。 +# 滑动窗口的内层循环结束时，右端点固定在 right，左端点在 0,1,2,…,left−1 的所有子串都是合法的，这一共有 left 个，加入答案。 +class Solution: + def numberOfSubstrings(self, s: str) -> int: + ans = left = 0 + cnt = defaultdict(int) + for c in s: + cnt[c] += 1 + while len(cnt) == 3: + out = s[left] # 离开窗口的字母 + cnt[out] -= 1 + if cnt[out] == 0: + del cnt[out] + left += 1 + ans += left + return ans +``` + +```python +#R 395 Longest Substring with At Least K Repeating Characters/至少有 K 个重复字符的最长子串 (Medium) s = "ababbc", k = 2 最长子串为 "ababb" +class Solution: + def longestSubstring(self, s: str, k: int) -> int: + ans = 0 + for i in range(1,27): #分治：遍历可能的字符种类数量（1~26） + cnt = Counter() #计数 + unique = 0 #字符种类 + num_k = 0 #满足出现次数大于等于k的字符串数量 + left = 0 + for right,char in enumerate(s): + if cnt[char] == 0: + unique += 1 + cnt[char] += 1 + if cnt[char] == k: + num_k += 1 + #当字符串种类超过i时，移动左窗口 + while unique > i: + left_char = s[left] + if cnt[left_char] == k: + num_k -= 1 #因为要一直移动左窗口，所以计数会一直减少，当进入循环时刚好==k时，再减一后，num_k就要减一了 + cnt[left_char] -= 1 + if cnt[left_char] == 0: #如果减完了，种类就少一个 + unique -= 1 + left += 1 + if unique == i and num_k == i: #当种类满足要求，且子串里的数量都满足>=k时，就更新ans + ans = max(ans ,right - left +1) + return ans +``` + +```python +#R 424 Longest Repeating Character Replacement/替换后的最长重复字符 (Medium) 可替换字符k次 +class Solution: + def characterReplacement(self, s: str, k: int) -> int: + count = [0 for _ in range(26)] #记录当前窗口的字母出现次数 + left = 0 #滑动窗口左边界 + right = 0 #滑动窗口右边界 + retval = 0 #最长窗口长度 + while right < len(s): + count[ord(s[right])-ord('A')] += 1 + benchmark = max(count) #选择出现次数最多的字母为基准 + others = sum(count) - benchmark #则其他字母需要通过替换操作来变为基准 + if others <= k: #通过与K进行比较来判断窗口是进行扩张？ + right += 1 + retval = max(retval, right-left)#记录当前有效窗口长度 + else: #通过与K进行比较来判断窗口还是进行位移？ + count[ord(s[left])-ord('A')] -= 1 + left += 1 + right += 1 #这里注意：位移操作需要整个向右移，不仅仅只是left向右 + return retval #返回最长窗口长度 +``` + +```python +#H LeetCode 33 "Search in Rotated Sorted Array" +def search(nums, target): + left, right = 0, len(nums) - 1 + while left <= right: + mid = (left + right) // 2 + if nums[mid] == target: + return mid + # 判断哪一部分是有序的 + if nums[left] <= nums[mid]: # 左半段有序 + if nums[left] <= target < nums[mid]: + right = mid - 1 + else: + left = mid + 1 + else: # 右半段有序 + if nums[mid] < target <= nums[right]: + left = mid + 1 + else: + right = mid - 1 + return -1 # 没有找到目标值 +``` + +```python +#H 81 Search in Rotated Sorted Array II/搜索旋转排序数组 II (Medium) +# 数组中的值不必互不相同,基于 33 题的简洁写法,只需增加一个 if +class Solution: + def search(self, nums: List[int], target: int) -> bool: + if not nums: + return False + n = len(nums) + if n == 1: + return nums[0] == target + + l, r = 0, n - 1 + while l <= r: + mid = (l + r) // 2 + if nums[mid] == target: + return True + if nums[l] == nums[mid] and nums[mid] == nums[r]: + l += 1 + r -= 1 + elif nums[l] <= nums[mid]: + if nums[l] <= target and target < nums[mid]: + r = mid - 1 + else: + l = mid + 1 + else: + if nums[mid] < target and target <= nums[n - 1]: + l = mid + 1 + else: + r = mid - 1 + return False +``` + +```python +#R 378 Kth Smallest Element in a Sorted Matrix/有序矩阵中第 K 小的元素 (Medium) +class Solution: + def kthSmallest(self, matrix: List[List[int]], k: int) -> int: + n = len(matrix) + def check(mid): + """遍历获取较小元素部分元素总数，并与k值比较""" + i, j = n-1, 0 + num = 0 + while i >= 0 and j < n: + if matrix[i][j] <= mid: + # 当前元素小于mid，则此元素及上方元素均小于mid + num += i + 1 + # 向右移动 + j += 1 + else: + # 当前元素大于mid，则向上移动，直到找到比mid小的值，或者出矩阵 + i -= 1 + return num >= k + left, right = matrix[0][0], matrix[-1][-1] + while left < right: + mid = (left + right) >> 1 + if check(mid): + # 满足 num >= k，范围太大，移动right至mid， 范围收缩 + right = mid + else: + left = mid + 1 + return left +``` + +```python +#R 410 Split Array Largest Sum/分割数组的最大值 (Hard) +# 分成 k 个非空的连续子数组，使得这 k 个子数组各自和的最大值 最小。 +class Solution: + def splitArray(self, nums: List[int], k: int) -> int: + def check(mx): + s, cnt = inf, 0 + for x in nums: + s += x + if s > mx: + s = x + cnt += 1 + return cnt <= k + left, right = max(nums), sum(nums) + return left + bisect_left(range(left, right + 1), True, key=check) +``` + +```python +#R 中位数可以奇偶统一表示 +def get_median_universal(nums): + sorted_nums = sorted(nums) + n = len(sorted_nums) + median = (sorted_nums[(n - 1) // 2] + sorted_nums[n // 2]) / 2.0 + return median +# Median of Two Sorted Arrays +import bisect +class Solution: + def findMedianSortedArrays(self, nums1: List[int], nums2: List[int]) -> float: + n = len(nums1) + len(nums2) + n1, n2 = len(nums1), len(nums2) + if n1 == 0: return (nums2[n2//2] + nums2[(n2-1)//2]) / 2 + if n2 == 0: return (nums1[n1//2] + nums1[(n1-1)//2]) / 2 + minN, maxN = min(nums1[0], nums2[0]), max(nums1[-1], nums2[-1]) + nums = list(range(minN, maxN+1)) + def func(x): + # nums1和nums2中<=x的数的个数（记为y），和x单调 + k1 = bisect.bisect_right(nums1, x) + k2 = bisect.bisect_right(nums2, x) + return k1 + k2 + k1 = bisect_right(nums, n//2, key=func) + if n % 2: k2 = k1 + else: k2 = bisect_right(nums, (n-1)//2, key=func) + return (nums[k1] + nums[k2]) / 2 +``` + +```python +#R 9 回文数 +class Solution: + def isPalindrome(self, x: int) -> bool: + # 同样地，如果数字的最后一位是 0，为了使该数字为回文，则其第一位数字也应该是 0 + if x < 0 or (x % 10 == 0 and x != 0): + return False + reverted_number = 0 + while x > reverted_number: + reverted_number = reverted_number * 10 + x % 10 + x //= 10 + # 当数字长度为奇数时，我们可以通过 reverted_number // 10 去除处于中位的数字。 + # 例如，当输入为 12321 时，在 while 循环的末尾我们可以得到 x = 12，reverted_number = 123， + # 由于处于中位的数字不影响回文（它总是与自己相等），所以我们可以简单地将其去除。 + return x == reverted_number or x == reverted_number // 10 +``` + +```python +#R 28. 实现 strStr()，next数组：最长的相同的真前后缀长度 +class Solution: + def getNext(self, next, s): + j = -1 + next[0] = j + for i in range(1, len(s)): + while j >= 0 and s[i] != s[j+1]: + j = next[j] + if s[i] == s[j+1]: + j += 1 + next[i] = j + def strStr(self, haystack: str, needle: str) -> int: + if not needle: + return 0 + next = [0] * len(needle) + self.getNext(next, needle) + j = -1 + for i in range(len(haystack)): + while j >= 0 and haystack[i] != needle[j+1]: + j = next[j] + if haystack[i] == needle[j+1]: + j += 1 + if j == len(needle) - 1: + return i - len(needle) + 1 + return -1 +``` + +```python +#R 767 Reorganize String/重构字符串 (Medium) +# 重新排布其中的字母，使得两相邻的字符不同。 +# 对于每一种元素，循环在不同桶中进行填充，由于桶的个数等于字符串中最多的元素的数目，因此每个桶中不会出现相同的元素，填充完毕后将桶依次相连即为答案 +class Solution: + def reorganizeString(self, s: str) -> str: + cnt, idx = Counter(s), 0 + bucketNum = cnt.most_common(1)[0][1] # 桶的数目等于字符串中最多的元素的数目 + if bucketNum > (len(s) + 1) // 2: return "" + buckets = [[] for _ in range(bucketNum)] + for c, num in cnt.most_common(): + for _ in range(num): + buckets[idx].append(c) + idx = (idx + 1) % bucketNum # 循环在不同桶中进行填充 + return "".join(["".join(bucket) for bucket in buckets]) +``` + +```python +#R 1053.9 Distant Barcodes/距离相等的条形码 (Medium) +# 请你重新排列这些条形码，使其中任意两个相邻的条形码不能相等。 +# 我们先统计数组barcodes中各个数出现的次数，然后按照它们出现的次数从大到小排序，依次填入偶数后奇数 +class Solution: + def rearrangeBarcodes(self, barcodes: List[int]) -> List[int]: + cnt = Counter(barcodes) + barcodes.sort(key=lambda x: (-cnt[x], x)) + n = len(barcodes) + ans = [0] * len(barcodes) + ans[::2] = barcodes[: (n + 1) // 2] + ans[1::2] = barcodes[(n + 1) // 2:] + return ans +``` + +```python +#R 92 Reverse Linked List II/反转链表 II (Medium) +class Solution: + def reverseBetween(self, head: Optional[ListNode], left: int, right: int) -> Optional[ListNode]: + p0 = dummy = ListNode(next=head) + for _ in range(left - 1): + p0 = p0.next + pre = None + cur = p0.next + for _ in range(right - left + 1): + nxt = cur.next + cur.next = pre # 每次循环只修改一个 next，方便大家理解 + pre = cur + cur = nxt + p0.next.next = cur + p0.next = pre + return dummy.next +``` + +```python +#R 146 LRU Cache/LRU 缓存机制 (Medium) +# get 如果关键字 key 存在于缓存中，则返回关键字的值，否则 -1 。 +# put 如果关键字 key 已经存在，则变更其数据值 value ；如果不存在，则向缓存中插入该组 key-value 。如果插入操作导致关键字数量超过 capacity ，则应该逐出最久未使用的关键字。 +class Node: + # 提高访问属性的速度，并节省内存 + __slots__ = 'prev', 'next', 'key', 'value' + def __init__(self, key=0, value=0): + self.key = key + self.value = value + self.prev = None + self.next = None +class LRUCache: + def __init__(self, capacity: int): + self.capacity = capacity + self.dummy = Node() # 哨兵节点 dummy.prev指向尾部，dummy.next指向头部 + self.dummy.prev = self.dummy + self.dummy.next = self.dummy + self.key_to_node = {} + # 获取 key 对应的节点，同时把该节点移到链表头部 + def get_node(self, key: int) -> Optional[Node]: + if key not in self.key_to_node: # 没有这本书 + return None + node = self.key_to_node[key] # 有这本书 + self.remove(node) # 把这本书抽出来 + self.push_front(node) # 放在最上面 + return node + def get(self, key: int) -> int: + # 返回关键字的值 + node = self.get_node(key) # get_node 会把对应节点移到链表头部 + return node.value if node else -1 + def put(self, key: int, value: int) -> None: + # 插入key-value 。如果关键字数量超过 capacity ，则应该逐出最久未使用的关键字 + node = self.get_node(key) + if node: # 有这本书 + node.value = value # 更新 value + return + self.key_to_node[key] = node = Node(key, value) # 新书 + self.push_front(node) # 放在最上面 + if len(self.key_to_node) > self.capacity: # 书太多了 + back_node = self.dummy.prev + del self.key_to_node[back_node.key] + self.remove(back_node) # 去掉最后一本书 + # 删除一个节点（抽出一本书） + def remove(self, x: Node) -> None: + x.prev.next = x.next + x.next.prev = x.prev + # 在链表头添加一个节点（把一本书放在最上面） + def push_front(self, x: Node) -> None: + x.prev = self.dummy + x.next = self.dummy.next + x.prev.next = x + x.next.prev = x +``` + +```python +#R 0 Decode String/字符串解码 (Medium) "3[a]2[bc]"，输出："aaabcbc" +class Solution: + def decodeString(self, s: str) -> str: + stack, res, multi = [], "", 0 + for c in s: + if c == '[': + stack.append([multi, res]) + res, multi = "", 0 + elif c == ']': + cur_multi, last_res = stack.pop() + res = last_res + cur_multi * res + elif '0' <= c <= '9': + multi = multi * 10 + int(c) + else: + res += c + return res +``` + +```python +#R 227 Basic Calculator II/基本计算器 II (Medium) +# 输入：s = "3+2*2" 输出：7。 当前元素为符号时，更新完栈后要记得更新数字和符号 +class Solution: + def calculate(self, s: str) -> int: + stack = [] + num = 0; sign = '+' + for i in range(len(s)): + if s[i].isdigit(): + num = num*10 + int(s[i]) + if s[i] in '+-*/' or i == len(s)-1: + if sign == '+': + stack.append(num) + elif sign == '-': + stack.append(-num) + elif sign == '*': + stack.append(stack.pop() * num) + else: + stack.append(int(stack.pop() / num)) + num = 0; sign = s[i] + return sum(stack) +``` + +```python +#H 385 Mini Parser/迷你语法分析器 (Medium) +# [123,[456,[789]]] [1,[2]] +class Solution: + def deserialize(self, s: str) -> NestedInteger: + if s[0] != '[': + return NestedInteger(int(s)) + stk = [] + num, negative = 0, False + for i in range(len(s)): + if s[i] == '[': + stk.append(NestedInteger()) + elif s[i] == '-': + negative = True + elif s[i].isdigit(): + num = num * 10 + int(s[i]) + elif s[i] == ',' or s[i] == ']': + if s[i - 1].isdigit(): + if negative: + num *= -1 + stk[-1].add(NestedInteger(num)) + num, negative = 0, False + if s[i] == ']' and len(stk) > 1: + ni = stk.pop() + stk[-1].add(ni) + return stk.pop() +``` + + + +```python +#R 636 Exclusive Time of Functions/函数的独占时间 (Medium) +# 当函数调用开始时，如果当前有函数正在运行，则当前正在运行函数应当停止，此时计算其的执行时间，然后将调用函数入栈。 +# 当函数调用结束时，将栈顶元素弹出，并计算相应的执行时间，如果此时栈顶有被暂停的函数，则开始运行该函数。 +# 输入：n = 2, logs = ["0:start:0","1:start:2","1:end:5","0:end:6"] 输出：两个函数的独占时间分别为[3,4] +class Solution: + def exclusiveTime(self, n: int, logs: List[str]) -> List[int]: + ans = [0] * n + st = [] + for log in logs: + idx, tp, timestamp = log.split(':') + idx, timestamp = int(idx), int(timestamp) + if tp[0] == 's': + if st: + ans[st[-1][0]] += timestamp - st[-1][1] + st[-1][1] = timestamp + st.append([idx, timestamp]) + else: + i, t = st.pop() + ans[i] += timestamp - t + 1 + if st: + st[-1][1] = timestamp + 1 + return ans +``` + + +```python +#R 921 Minimum Add to Make Parentheses Valid/使括号有效的最少添加 (Medium) +class Solution: + def minAddToMakeValid(self, s: str) -> int: + stk = [] + for c in s: + if c == ')' and stk and stk[-1] == '(': + stk.pop() + else: + stk.append(c) + return len(stk) +``` + +# 单调栈 +https://blog.csdn.net/zy_dreamer/article/details/131036101 +从左到右遍历元素。单调递增栈：栈顶最小。 +- 查找 「比当前元素大的元素」 就用 单调递增栈，查找 「比当前元素小的元素」 就用 单调递减栈。 +- 从 「左侧」 查找就看 「插入栈」 时的栈顶元素，从 「右侧」 查找就看 「弹出栈」 时即将插入的元素。 + +```python +#H 模板，单调递增栈（递减改为小于等于） +def monotoneIncreasingStack(nums): + stack = [] + for num in nums: + while stack and num >= stack[-1]: + stack.pop() + stack.append(num) +``` + + +```python +#R 0 每日温度，观测到更高的气温，至少需要等待的天数 +class Solution: + def dailyTemperatures(self, T: List[int]) -> List[int]: + n = len(T) + stack = [] + ans = [0 for _ in range(n)] + for i in range(n): + while stack and T[i] > T[stack[-1]]: + index = stack.pop() + ans[index] = (i-index) + stack.append(i) + return ans +``` + + +```python +#H 0 接雨水，排列的柱子，下雨之后能接多少雨水。 +# 方法总结：找上一个更大元素，在找的过程中填坑。while中加了等号，这可以让栈中没有重复元素，以节省空间。 https://www.bilibili.com/video/BV1VN411J7S7/?vd_source=3f76269c2962e01f0d61bbeac282e5d2 单调递减栈。 +class Solution: + def trap(self, height: List[int]) -> int: + ans = 0 + st = [] + for i, h in enumerate(height): + while st and h >= height[st[-1]]: + bottom_h = height[st.pop()] + if not st: # len(st) == 0 + break + left = st[-1] + dh = min(height[left], h) - bottom_h # 面积的高 + ans += dh * (i - left - 1) + st.append(i) + return ans +``` + + + +```python +#R 0.下一个更大元素 I, 每一nums1中的x在nums2中对应位置右侧的第一个比 x 大的元素。如果不存在 -1 +class Solution: + def nextGreaterElement(self, nums1: List[int], nums2: List[int]) -> List[int]: + res = {} + stack = [] + for num in reversed(nums2): + while stack and num >= stack[-1]: + stack.pop() + res[num] = stack[-1] if stack else -1 + stack.append(num) + return [res[num] for num in nums1] +``` + + +```python +#H 0 柱状图中最大的矩形。n个非负整数，用来表示柱状图中各个柱子的高度。求该柱状图中勾勒出来的矩形的最大面积。 +# 在i左侧的小于h的最近元素的下标left，右侧的小于h的最近元素的下标right，矩形的宽度就是right−left−1 +class Solution: + def largestRectangleArea(self, heights: List[int]) -> int: + n, heights, st, ans = len(heights), [0] + heights + [0], [], 0 + for i in range(n + 2): + while st and heights[st[-1]] > heights[i]: + ans = max(ans, heights[st.pop(-1)] * (i - st[-1] - 1)) + st.append(i) + return ans +``` + +```python +#H 85 Maximal Rectangle/最大矩形 (Hard) +class Solution: + def maximalRectangle(self, matrix: List[List[str]]) -> int: + m = len(matrix) + if m == 0: return 0 + n = len(matrix[0]) + heights = [0] * n + ans = 0 + for i in range(m): + for j in range(n): + if matrix[i][j] == "0": + heights[j] = 0 + else: + heights[j] += 1 + ans = max(ans, self.largestRectangleArea(heights)) + return ans +``` + + +```python +#H 315. 计算右侧小于当前元素的个数 困难 +# counts[i] 的值是 nums[i] 右侧小于 nums[i] 的元素的数量。输入数组反过来插入一个有序数组（降序）中，插入的位置就是在原数组中位于它右侧的元素的个数。 +class Solution: + def countSmaller(self, nums: List[int]) -> List[int]: + sortns = [] + res = [] + for n in reversed(nums): + idx = bisect.bisect_left(sortns, n) + res.append(idx) + sortns.insert(idx,n) + return res[::-1] +``` + +```python +#R 找前面比自己小的元素 +def find_smaller_elements(nums): + result = [] + stack = [] # 用于存储元素的值 + for num in nums: + while stack and stack[-1] >= num: + stack.pop() + if not stack: + result.append(None) # 如果前面没有比当前元素小的元素，可以用 None 表示 + else: + result.append(stack[-1]) + stack.append(num) + return result +``` + + +```python +#R 503 Next Greater Element II/下一个更大元素 II (Medium) +class Solution: + def nextGreaterElements(self, nums: List[int]) -> List[int]: + n = len(nums) + ans = [-1] * n + stack = [] + for i in range(n * 2 - 1): + while stack and nums[i % n] > nums[stack[-1]]: + ans[stack.pop()] = nums[i % n] + stack.append(i % n) + return ans +``` + +```python +#R 768 Max Chunks To Make Sorted II/最多能完成排序的块 II (Hard) +# 分割成若干块 ，分别进行排序。之后再连接起来，使得连接的结果和按升序排序后的原数组相同。 +# 从左到右，每个分块都有一个最大值，并且这些分块的最大值呈单调递增（非严格递增）321 65 +class Solution: + def maxChunksToSorted(self, arr: List[int]) -> int: + stk = [] + for v in arr: + if not stk or v >= stk[-1]: + stk.append(v) + else: + mx = stk.pop() + while stk and stk[-1] > v: + stk.pop() + stk.append(mx) + return len(stk) +``` + + + +# 堆 + + +```python +#R 502 IPO (Hard) +# 贪心 + 排序 + 大顶堆：选择当前可启动的最大利润项目 +# 利润profits，启动该项目需要的最小资本capital。最初你的资本为w,选择最多k个项目 +class Solution: + def findMaximizedCapital(self, k: int, w: int, profits: List[int], capital: List[int]) -> int: + # 生成索引序列 并 根据资本值对索引进行升序排序 + n = len(capital) + indexes = sorted( [i for i in range(n)], key=lambda i: capital[i]) + pq = [] # 维护堆内利润值的大顶堆 + i = 0 + while k > 0: + # 将启动资本小于等于当前资本的项目的利润加入大顶堆 + while i < n and capital[indexes[i]] <= w: + heapq.heappush(pq, -profits[indexes[i]]) # 取相反数实现大顶堆 + i += 1 + if not pq: break # 没有可以启动的项目，后面启动资本更大的项目也无法启动，退出 + w += -heappop(pq) # 选择启动资本满足条件的项目中利润最大的那个，更新w + k -= 1 + return w +``` + +```python +#R 快速排序 +def quick_sort(arr, low=0, high=None): + if high is None: + high = len(arr) - 1 + + def partition(arr, low, high): + pivot = arr[high] + i = low - 1 + for j in range(low, high): + # 如果当前元素小于等于基准 + if arr[j] <= pivot: + # 交换元素 + i += 1 + arr[i], arr[j] = arr[j], arr[i] + # 将基准放到正确位置 + arr[i + 1], arr[high] = arr[high], arr[i + 1] + return i + 1 + + def _quick_sort(arr, low, high): + if low < high: + partition_index = partition(arr, low, high) + _quick_sort(arr, low, partition_index - 1) + _quick_sort(arr, partition_index + 1, high) + + _quick_sort(arr, low, high) + return arr +``` + + +```python +#R 堆排序 https://blog.csdn.net/Solititude/article/details/129182217 +def heapify(arr, n, i): + largest = i + left_child = 2 * i + 1 + right_child = 2 * i + 2 + if left_child < n and arr[left_child] > arr[largest]: + largest = left_child + if right_child < n and arr[right_child] > arr[largest]: + largest = right_child + if largest != i: + arr[i], arr[largest] = arr[largest], arr[i] + heapify(arr, n, largest) +def heap_sort(arr): + n = len(arr) + # 构建最大堆 + for i in range(n // 2 - 1, -1, -1): + heapify(arr, n, i) + # 逐步取出堆顶元素，进行堆排序 + for i in range(n - 1, 0, -1): + arr[0], arr[i] = arr[i], arr[0] + heapify(arr, i, 0) +``` + + +```python +#R 709.99 Random Pick with Blacklist/黑名单中的随机数 (Hard) +# 选取一个 未加入 黑名单 blacklist 的整数 +# 构建一个从 [0,n−m) 范围内的黑名单数到 [n−m,n) 的白名单数的映射 +class Solution: + def __init__(self, n: int, blacklist: List[int]): + m = len(blacklist) + self.bound = w = n - m + black = {b for b in blacklist if b >= self.bound} + self.b2w = {} + for b in blacklist: + if b < self.bound: + while w in black: + w += 1 + self.b2w[b] = w + w += 1 + def pick(self) -> int: + x = randrange(self.bound) + return self.b2w.get(x, x) +``` + +# 累加和 + + +```python +#R 862. Shortest Subarray with Sum at Least K +import heapq +def shortestSubarray(nums, k): + # 初始化结果为正无穷大，累加和为0，优先队列用于保存累加和及其对应的下标 + res = float('inf') + sum_val = 0 + pq = [(0, -1)] + for i in range(len(nums)): + # 计算当前位置的累加和 + sum_val += nums[i] + # 检查队列中是否有满足条件的累加和，如果有，更新结果 + while pq and sum_val - pq[0][0] >= k: + res = min(res, i - pq[0][1]) + heapq.heappop(pq) + # 将当前累加和及其下标加入队列 + heapq.heappush(pq, (sum_val, i)) + # 如果结果仍然是正无穷大，说明没有符合条件的子数组 + return -1 if res == float('inf') else res +``` + + +```python +#H 负二进制加法/Adding Two Negabinary Numbers +# 如果 x≥2，那么将 x 减去 2，并向高位进位 −1。即逢 2 进负 1。 +# 如果 x=−1，由于 −(−2)^i=(−2)^i +(−2)^{i+1}，所以我们可以将 x 置为 1，并向高位进位 1。 +class Solution: + def addNegabinary(self, arr1: List[int], arr2: List[int]) -> List[int]: + i, j = len(arr1) - 1, len(arr2) - 1 + c = 0 + ans = [] + while i >= 0 or j >= 0 or c: + a = 0 if i < 0 else arr1[i] + b = 0 if j < 0 else arr2[j] + x = a + b + c + c = 0 + if x >= 2: + x -= 2 + c -= 1 + elif x == -1: + x = 1 + c += 1 + ans.append(x) + i, j = i - 1, j - 1 + while len(ans) > 1 and ans[-1] == 0: + ans.pop() + return ans[::-1] +``` + + +```python +#R 二叉树的迭代遍历 +# 前序遍历-迭代-LC144_二叉树的前序遍历 +class Solution: + def preorderTraversal(self, root: TreeNode) -> List[int]: + # 根结点为空则返回空列表 + if not root: + return [] + stack = [root] + result = [] + while stack: + node = stack.pop() + # 中结点先处理 + result.append(node.val) + # 右孩子先入栈 + if node.right: + stack.append(node.right) + # 左孩子后入栈 + if node.left: + stack.append(node.left) + return result +``` + + +```python +#R 中序遍历-迭代-LC94_二叉树的中序遍历 +class Solution: + def inorderTraversal(self, root: TreeNode) -> List[int]: + if not root: + return [] + stack = [] # 不能提前将root结点加入stack中 + result = [] + cur = root + while cur or stack: + # 先迭代访问最底层的左子树结点 + if cur: + stack.append(cur) + cur = cur.left + # 到达最左结点后处理栈顶结点 + else: + cur = stack.pop() + result.append(cur.val) + # 取栈顶元素右结点 + cur = cur.right + return result +``` + + + +```python +#R 102.二叉树的层序遍历 +# 利用长度法 +# Definition for a binary tree node. +# class TreeNode: +# def __init__(self, val=0, left=None, right=None): +# self.val = val +# self.left = left +# self.right = right +class Solution: + def levelOrder(self, root: Optional[TreeNode]) -> List[List[int]]: + if not root: + return [] + queue = collections.deque([root]) + result = [] + while queue: + level = [] + for _ in range(len(queue)): + cur = queue.popleft() + level.append(cur.val) + if cur.left: + queue.append(cur.left) + if cur.right: + queue.append(cur.right) + result.append(level) + return result +``` + + + +```python +#R 671. 二叉树中第二小的节点。 +# 给定一个非空特殊的二叉树，每个节点都是正数，并且每个节点的子节点数量只能为 2 或 0。如果一个节点有两个子节点的话，那么该节点的值等于两个子节点中较小的一个。 +class Solution: + def findSecondMinimumValue(self, root: TreeNode) -> int: + if not root or not root.left: + return -1 + # 我们知道root.val是最小值，那么 + # 第二小的值存在于 更小的子节点那一边的子树的第二小的值 或 更大的子节点 之中 + left = root.left.val if root.left.val != root.val else self.findSecondMinimumValue(root.left) + right = root.right.val if root.right.val != root.val else self.findSecondMinimumValue(root.right) + return min(left, right) if left != -1 and right != -1 else max(left, right) +``` + + + +```python +#R 208 前缀树是一种特殊的多叉树，它的 TrieNode 中 chidren 是一个大小为 26 的一维数组，分别对应了26个英文字符 void insert(String word) 向前缀树中插入字符串 word，search(String word) 如果字符串 word 在前缀树中，返回 true（即，在检索之前已经插入）；否则，返回 false 。 startsWith 如果之前已经插入的字符串 word 的前缀之一为 prefix ，返回 true ；否则，返回 false 。 +class Trie: + def __init__(self): + self.children = [None] * 26 + self.isEnd = False + def searchPrefix(self, prefix: str) -> "Trie": + node = self + for ch in prefix: + ch = ord(ch) - ord("a") + if not node.children[ch]: + return None + node = node.children[ch] + return node + def insert(self, word: str) -> None: + node = self + for ch in word: + ch = ord(ch) - ord("a") + if not node.children[ch]: + node.children[ch] = Trie() + node = node.children[ch] + node.isEnd = True + def search(self, word: str) -> bool: + node = self.searchPrefix(word) + return node is not None and node.isEnd + def startsWith(self, prefix: str) -> bool: + return self.searchPrefix(prefix) is not None +``` + + +```python +#R 1032. Stream of Characters 接收一个字符流，并检查这些字符的后缀是否是字符串数组 words 中的一个字符串。 +class Trie: + def __init__(self): + self.children = [None] * 26 + self.is_end = False + def insert(self, w: str): + node = self + for c in w[::-1]: + idx = ord(c) - ord('a') + if node.children[idx] is None: + node.children[idx] = Trie() + node = node.children[idx] + node.is_end = True + def search(self, w: List[str]) -> bool: + node = self + for c in w[::-1]: + idx = ord(c) - ord('a') + if node.children[idx] is None: + return False + node = node.children[idx] + if node.is_end: + return True + return False +class StreamChecker: + def __init__(self, words: List[str]): + self.trie = Trie() + self.cs = [] + self.limit = 201 + for w in words: + self.trie.insert(w) + def query(self, letter: str) -> bool: + self.cs.append(letter) + return self.trie.search(self.cs[-self.limit:]) +# Your StreamChecker object will be instantiated and called as such: +# obj = StreamChecker(words) +# param_1 = obj.query(letter) +``` + + + + + + +```python +#R 124 Binary Tree Maximum Path Sum/二叉树中的最大路径和 (Hard) +# 输入：root = [1,2,3], 最优路径是 2 -> 1 -> 3 +# 链：从下面的某个节点（不一定是叶子）到当前节点的路径。把这条链的节点值之和，作为 dfs 的返回值。如果节点值之和是负数，则返回 0。 +# 直径：等价于由两条（或者一条）链拼成的路径。我们枚举每个 node，假设直径在这里「拐弯」，也就是计算由左右两条从下面的某个节点（不一定是叶子）到 node 的链的节点值之和，去更新答案的最大值。 +# dfs 返回的是链的节点值之和，不是直径的节点值之和。 +class Solution: + def maxPathSum(self, root: Optional[TreeNode]) -> int: + ans = -inf + def dfs(node: Optional[TreeNode]) -> int: + if node is None: + return 0 # 没有节点，和为 0 + l_val = dfs(node.left) # 左子树最大链和 + r_val = dfs(node.right) # 右子树最大链和 + nonlocal ans + ans = max(ans, l_val + r_val + node.val) # 两条链拼成路径 + return max(max(l_val, r_val) + node.val, 0) # 当前子树最大链和（注意这里和 0 取最大值了） + dfs(root) + return ans +``` + +```python +#R 297 Serialize and Deserialize Binary Tree/二叉树的序列化与反序列化 (Medium) +# 序列化 使用层序遍历实现。反序列化通过递推公式反推各节点在序列中的索引 +class Codec: + def serialize(self, root): + if not root: return "[]" + queue = collections.deque() + queue.append(root) + res = [] + while queue: + node = queue.popleft() + if node: + res.append(str(node.val)) + queue.append(node.left) + queue.append(node.right) + else: res.append("null") + return '[' + ','.join(res) + ']' + + def deserialize(self, data): + if data == "[]": return + vals, i = data[1:-1].split(','), 1 + root = TreeNode(int(vals[0])) + queue = collections.deque() + queue.append(root) + while queue: + node = queue.popleft() + if vals[i] != "null": + node.left = TreeNode(int(vals[i])) + queue.append(node.left) + i += 1 + if vals[i] != "null": + node.right = TreeNode(int(vals[i])) + queue.append(node.right) + i += 1 + return root +``` + +```python +#R 366 Find Leaves of Binary Tree $/寻找二叉树的叶子节点 (Medium) +# 收集所有的叶子节点；移除所有的叶子节点；重复以上步骤，直到树为空。 高度代表倒数第几个叶子节点 +class Solution: + def findLeaves(self, root: TreeNode) -> List[List[int]]: + # 自底向上递归 + def dfs(root): + if not root:return 0 + l,r=dfs(root.left),dfs(root.right) + depth=max(l,r)+1 + res[depth].append(root.val) + return depth + + res=collections.defaultdict(list) + dfs(root) + return [v for v in res.values()] +``` + +```python +#R 652 Find Duplicate Subtrees/寻找重复的子树 (Medium) +# 递归的方法进行序列化 +class Solution: + def findDuplicateSubtrees(self, root: Optional[TreeNode]) -> List[Optional[TreeNode]]: + def dfs(node: Optional[TreeNode]) -> str: + if not node: + return "" + serial = "".join([str(node.val), "(", dfs(node.left), ")(", dfs(node.right), ")"]) + if (tree := seen.get(serial, None)): + repeat.add(tree) + else: + seen[serial] = node + return serial + seen = dict() + repeat = set() + dfs(root) + return list(repeat) +``` + +```python +#R 0 My Calendar II/我的日程安排表 II (Medium) +# 如果要添加的时间内不会导致三重预订时，则可以存储这个新的日程安排。 +class MyCalendarTwo: + def __init__(self): + self.booked = [] + self.overlaps = [] + def book(self, start: int, end: int) -> bool: + if any(s < end and start < e for s, e in self.overlaps): + return False + # 注意两个区间不相交的补=s < end and start < e + for s, e in self.booked: + if s < end and start < e: + self.overlaps.append((max(s, start), min(e, end))) + self.booked.append((start, end)) + return True +``` + + +```python +#R 987 Vertical Order Traversal of a Binary Tree/二叉树的垂序遍历 (Medium) +class Solution: + def verticalTraversal(self, root: Optional[TreeNode]) -> List[List[int]]: + groups = defaultdict(list) + def dfs(node, row, col): #二叉树先序遍历 + if node is None: + return + groups[col].append((row, node.val)) # col 相同的分到同一组 + dfs(node.left, row + 1, col - 1) + dfs(node.right, row + 1, col + 1) + + dfs(root, 0, 0) + ans = [] + for _, g in sorted(groups.items()): + g.sort() # 按照 row 排序，row 相同按照 val 排序 + ans.append([val for _, val in g]) + return ans +``` + + +# 回溯算法 + +组合 + + +```python +#R 216.组合总和III +# [1,2,3,4,5,6,7,8,9]这个集合中找到和为n的k个数的组合 +class Solution: + def combinationSum3(self, k: int, n: int) -> List[List[int]]: + result = [] # 存放结果集 + self.backtracking(n, k, 0, 1, [], result) + return result + def backtracking(self, targetSum, k, currentSum, startIndex, path, result): + if currentSum > targetSum: # 剪枝操作 + return # 如果path的长度等于k但currentSum不等于targetSum，则直接返回 + if len(path) == k: + if currentSum == targetSum: + result.append(path[:]) + return + for i in range(startIndex, 9 - (k - len(path)) + 2): # 剪枝 + currentSum += i # 处理 + path.append(i) # 处理 + self.backtracking(targetSum, k, currentSum, i + 1, path, result) # 注意i+1调整startIndex + currentSum -= i # 回溯 + path.pop() # 回溯 +``` + + + +```python +#R 40.组合总和II +# 数组 candidates 和一个目标数 target ，找出 candidates 中所有可以使数字和为 target 的组合。candidates 中的每个数字在每个组合中只能使用一次。 +class Solution: + def backtracking(self, candidates, target, total, startIndex, used, path, result): + if total == target: + result.append(path[:]) + return + for i in range(startIndex, len(candidates)): + # 对于相同的数字，只选择第一个未被使用的数字，跳过其他相同数字 + if i > startIndex and candidates[i] == candidates[i - 1] and not used[i - 1]: + continue + if total + candidates[i] > target: + break + total += candidates[i] + path.append(candidates[i]) + used[i] = True + self.backtracking(candidates, target, total, i + 1, used, path, result) + used[i] = False + total -= candidates[i] + path.pop() + def combinationSum2(self, candidates, target): + used = [False] * len(candidates) + result = [] + candidates.sort() + self.backtracking(candidates, target, 0, 0, used, [], result) + return result +``` + + +分割 + + + +```python +#R 131.分割回文串 +# 将 s 分割成一些子串，使每个子串都是回文串 +class Solution: + def partition(self, s: str) -> List[List[str]]: + # 递归用于纵向遍历; for循环用于横向遍历; 当切割线迭代至字符串末尾，说明找到一种方法; 类似组合问题，为了不重复切割同一位置，需要start_index来做标记下一轮递归的起始位置(切割线) + result = [] + self.backtracking(s, 0, [], result) + return result + def backtracking(self, s, start_index, path, result ): + # Base Case + if start_index == len(s): + result.append(path[:]) + return + # 单层递归逻辑 + for i in range(start_index, len(s)): + # 此次比其他组合题目多了一步判断： + # 判断被截取的这一段子串([start_index, i])是否为回文串 + if self.is_palindrome(s, start_index, i): + path.append(s[start_index:i+1]) + self.backtracking(s, i+1, path, result) # 递归纵向遍历：从下一处进行切割，判断其余是否仍为回文串 + path.pop() # 回溯 + def is_palindrome(self, s: str, start: int, end: int) -> bool: + i: int = start + j: int = end + while i < j: + if s[i] != s[j]: + return False + i += 1 + j -= 1 + return True +``` + + +```python +#R 93.复原IP地址 +class Solution: + def restoreIpAddresses(self, s: str) -> List[str]: + result = [] + self.backtracking(s, 0, 0, "", result) + return result + def backtracking(self, s, start_index, point_num, current, result): + if point_num == 3: # 逗点数量为3时，分隔结束 + if self.is_valid(s, start_index, len(s) - 1): # 判断第四段子字符串是否合法 + current += s[start_index:] # 添加最后一段子字符串 + result.append(current) + return + for i in range(start_index, len(s)): + if self.is_valid(s, start_index, i): # 判断 [start_index, i] 这个区间的子串是否合法 + sub = s[start_index:i + 1] + self.backtracking(s, i + 1, point_num + 1, current + sub + '.', result) + else: + break + def is_valid(self, s, start, end): + if start > end: + return False + if s[start] == '0' and start != end: # 0开头的数字不合法 + return False + num = 0 + for i in range(start, end + 1): + if not s[i].isdigit(): # 遇到非数字字符不合法 + return False + num = num * 10 + int(s[i]) + if num > 255: # 如果大于255了不合法 + return False + return True +``` + +子集 + + +```python +#R 90.子集II +# 可能包含重复元素的整数数组 nums，返回该数组所有可能的子集（幂集）。 解集不能包含重复的子集。 +class Solution: + def subsetsWithDup(self, nums): + result = [] + path = [] + used = [False] * len(nums) + nums.sort() # 去重需要排序 + self.backtracking(nums, 0, used, path, result) + return result + def backtracking(self, nums, startIndex, used, path, result): + result.append(path[:]) # 收集子集 + for i in range(startIndex, len(nums)): + # used[i - 1] == True，说明同一树枝 nums[i - 1] 使用过 + # used[i - 1] == False，说明同一树层 nums[i - 1] 使用过 + # 而我们要对同一树层使用过的元素进行跳过 + if i > 0 and nums[i] == nums[i - 1] and not used[i - 1]: + continue + path.append(nums[i]) + used[i] = True + self.backtracking(nums, i + 1, used, path, result) + used[i] = False + path.pop() +``` + +```python +#R 491.递增子序列 +# 数组中可能包含重复数字，相等的数字应该被视为递增的一种情况。 利用set去重 +class Solution: + def findSubsequences(self, nums): + result = [] + path = [] + self.backtracking(nums, 0, path, result) + return result + def backtracking(self, nums, startIndex, path, result): + if len(path) > 1: + result.append(path[:]) # 注意要使用切片将当前路径的副本加入结果集 + # 注意这里不要加return，要取树上的节点 + + uset = set() # 使用集合对本层元素进行去重 + for i in range(startIndex, len(nums)): + if (path and nums[i] < path[-1]) or nums[i] in uset: + continue + + uset.add(nums[i]) # 记录这个元素在本层用过了，本层后面不能再用了 + path.append(nums[i]) + self.backtracking(nums, i + 1, path, result) + path.pop() +``` + +排列 + + +```python +#R 47.全排列 II +# 可包含重复数字的序列 ，返回所有不重复的全排列 +class Solution: + def permuteUnique(self, nums): + nums.sort() # 排序 + result = [] + self.backtracking(nums, [], [False] * len(nums), result) + return result + def backtracking(self, nums, path, used, result): + if len(path) == len(nums): + result.append(path[:]) + return + for i in range(len(nums)): + if (i > 0 and nums[i] == nums[i - 1] and not used[i - 1]) or used[i]: + continue + used[i] = True + path.append(nums[i]) + self.backtracking(nums, path, used, result) + path.pop() + used[i] = False +``` + + +棋盘 + +```python +#R 51. N皇后 +class Solution: + def solveNQueens(self, n: int) -> List[List[str]]: + result = [] # 存储最终结果的二维字符串数组 + chessboard = ['.' * n for _ in range(n)] # 初始化棋盘 + self.backtracking(n, 0, chessboard, result) # 回溯求解 + return [[''.join(row) for row in solution] for solution in result] # 返回结果集 + def backtracking(self, n: int, row: int, chessboard: List[str], result: List[List[str]]) -> None: + if row == n: + result.append(chessboard[:]) # 棋盘填满，将当前解加入结果集 + return + for col in range(n): + if self.isValid(row, col, chessboard): + chessboard[row] = chessboard[row][:col] + 'Q' + chessboard[row][col+1:] # 放置皇后 + self.backtracking(n, row + 1, chessboard, result) # 递归到下一行 + chessboard[row] = chessboard[row][:col] + '.' + chessboard[row][col+1:] # 回溯，撤销当前位置的皇后 + def isValid(self, row: int, col: int, chessboard: List[str]) -> bool: + # 检查列 + for i in range(row): + if chessboard[i][col] == 'Q': + return False # 当前列已经存在皇后，不合法 + # 检查 45 度角是否有皇后 + i, j = row - 1, col - 1 + while i >= 0 and j >= 0: + if chessboard[i][j] == 'Q': + return False # 左上方向已经存在皇后，不合法 + i -= 1 + j -= 1 + # 检查 135 度角是否有皇后 + i, j = row - 1, col + 1 + while i >= 0 and j < len(chessboard): + if chessboard[i][j] == 'Q': + return False # 右上方向已经存在皇后，不合法 + i -= 1 + j += 1 + return True # 当前位置合法 +``` + + +```python +#R 37. 解数独 +class Solution: + def solveSudoku(self, board: List[List[str]]) -> None: + """ + Do not return anything, modify board in-place instead. + """ + self.backtracking(board) + def backtracking(self, board: List[List[str]]) -> bool: + # 若有解，返回True；若无解，返回False + for i in range(len(board)): # 遍历行 + for j in range(len(board[0])): # 遍历列 + # 若空格内已有数字，跳过 + if board[i][j] != '.': continue + for k in range(1, 10): + if self.is_valid(i, j, k, board): + board[i][j] = str(k) + if self.backtracking(board): return True + board[i][j] = '.' + # 若数字1-9都不能成功填入空格，返回False无解 + return False + return True # 有解 + def is_valid(self, row: int, col: int, val: int, board: List[List[str]]) -> bool: + # 判断同一行是否冲突 + for i in range(9): + if board[row][i] == str(val): + return False + # 判断同一列是否冲突 + for j in range(9): + if board[j][col] == str(val): + return False + # 判断同一九宫格是否有冲突 + start_row = (row // 3) * 3 + start_col = (col // 3) * 3 + for i in range(start_row, start_row + 3): + for j in range(start_col, start_col + 3): + if board[i][j] == str(val): + return False + return True +``` + +其他 + + +```python +#R 282 Expression Add Operators/给表达式添加运算符 (Hard) +# 输入: num = "123", target = 6 输出: ["1+2+3", "1*2*3"] +# 每个字符与字符实际上有四种连接方式，加减乘和拼接 +class Solution: + def addOperators(self, num: str, target: int) -> List[str]: + def backtrace(idx: int, cursum: int, preadd: int) -> None: + nonlocal res + nonlocal path + if idx == n: + if cursum == target: + res.append(path[:]) + return + pn = len(path) + for i in range(idx, n): + x_str = num[idx: i + 1] + x = int(x_str) + if idx == 0: + path += x_str + backtrace(i + 1, cursum + x, x) + path = path[ :pn] + else: + path += '+' + x_str + backtrace(i + 1, cursum + x, x) + path = path[ :pn] + path += '-' + x_str + backtrace(i + 1, cursum - x, -x) + path = path[ :pn] + path += '*' + x_str + backtrace(i + 1, cursum - preadd + preadd * x, preadd * x) + path = path[ :pn] + if x == 0: + return + n = len(num) + res = [] + path = "" + backtrace(0, 0, 0) + return res +``` + +```python +#R 698 Partition to K Equal Sum Subsets/划分为k个相等的子集 (Medium) +class Solution: + def canPartitionKSubsets(self, nums: List[int], k: int) -> bool: + def dfs(i): + if i == len(nums): + return True + for j in range(k): + if j and cur[j] == cur[j - 1]: + continue + cur[j] += nums[i] + if cur[j] <= s and dfs(i + 1): + return True + cur[j] -= nums[i] + return False + s, mod = divmod(sum(nums), k) + if mod: + return False + cur = [0] * k + nums.sort(reverse=True) + return dfs(0) +``` + + +# 图Visited + + +```python +#R Most Stones Removed with Same Row or Column/移除最多的同行或同列石头(Medium) +# 石头放在整数坐标点上, move操作移除某一块石头共享一列或一行的一块石头,最多能执行多少次 move +class Solution: + def removeStones(self, stones: List[List[int]]) -> int: + # 构建图 + graph = collections.defaultdict(list) + for i, (x, y) in enumerate(stones): + for j, (xx, yy) in enumerate(stones): + if x == xx or y == yy: + graph[i].append(j) + # DFS计算连通分量数量 + def dfs(node): + visited.add(node) + for neighbor in graph[node]: + if neighbor not in visited: + dfs(neighbor) + n = len(stones) + visited = set() + components = 0 + for i in range(n): + if i not in visited: + components += 1 + dfs(i) + return n - components +``` + + + +```python +#R 200. Number of Islands 给定一个由 '1'（陆地）和 '0'（水）组成的二维网格，计算岛屿的数量。岛屿是由相邻的陆地水平或垂直连接形成的（不包括对角线）。你可以假设网格的四个边都被水包围。 +class Solution: + def numIslands(self, grid: List[List[str]]) -> int: + if not grid or not grid[0]: + return 0 + rows, cols = len(grid), len(grid[0]) + islands = 0 + def dfs(row, col): + if 0 <= row < rows and 0 <= col < cols and grid[row][col] == '1': + grid[row][col] = '0' # 标记为已访问 + # 递归搜索相邻的陆地 + dfs(row - 1, col) + dfs(row + 1, col) + dfs(row, col - 1) + dfs(row, col + 1) + for row in range(rows): + for col in range(cols): + if grid[row][col] == '1': + islands += 1 + dfs(row, col) + return islands +``` + + +```python +#R (hard) 单词接龙，从单词 beginWord 和 endWord 的转换序列每次转换只能改变一个字母，wordList = ["hot","dot","dog"] +class Solution: + def ladderLength(self, beginWord: str, endWord: str, wordList: List[str]) -> int: + wordSet = set(wordList) + if len(wordSet)== 0 or endWord not in wordSet: + return 0 + mapping = {beginWord:1} + queue = deque([beginWord]) + while queue: + word = queue.popleft() + path = mapping[word] + for i in range(len(word)): + word_list = list(word) + for j in range(26): + word_list[i] = chr(ord('a')+j) + newWord = "".join(word_list) + if newWord == endWord: + return path+1 + if newWord in wordSet and newWord not in mapping: + mapping[newWord] = path+1 + queue.append(newWord) + return 0 +``` + + +```python +#R 463. 岛屿的周长 +class Solution: + def islandPerimeter(self, grid: List[List[int]]) -> int: + m = len(grid) + n = len(grid[0]) + # 创建res二维素组记录答案 + res = [[0] * n for j in range(m)] + for i in range(m): + for j in range(len(grid[i])): + # 如果当前位置为水域，不做修改或reset res[i][j] = 0 + if grid[i][j] == 0: + res[i][j] = 0 + # 如果当前位置为陆地，往四个方向判断，update res[i][j] + elif grid[i][j] == 1: + if i == 0 or (i > 0 and grid[i-1][j] == 0): + res[i][j] += 1 + if j == 0 or (j >0 and grid[i][j-1] == 0): + res[i][j] += 1 + if i == m-1 or (i < m-1 and grid[i+1][j] == 0): + res[i][j] += 1 + if j == n-1 or (j < n-1 and grid[i][j+1] == 0): + res[i][j] += 1 + # 最后求和res矩阵，这里其实不一定需要矩阵记录，可以设置一个variable res 记录边长，舍矩阵无非是更加形象而已 + ans = sum([sum(row) for row in res]) + return ans +``` + + +```python +#R 1091. Shortest Path in Binary Matrix/二进制矩阵中的最短路径(Medium) +class Solution: + def shortestPathBinaryMatrix(self, grid: List[List[int]]) -> int: + if grid[0][0]: + return -1 + n = len(grid) + grid[0][0] = 1 + q = deque([(0, 0)]) + ans = 1 + while q: + for _ in range(len(q)): + i, j = q.popleft() + if i == j == n - 1: + return ans + for x in range(i - 1, i + 2): + for y in range(j - 1, j + 2): + if 0 <= x < n and 0 <= y < n and grid[x][y] == 0: + grid[x][y] = 1 + q.append((x, y)) + ans += 1 + return -1 +``` + +```python +#R 210 Course Schedule II/课程表 II (Medium) +# 返回你为了学完所有课程所安排的学习顺序，indeg 存储每个节点的入度 +class Solution: + def findOrder(self, numCourses: int, prerequisites: List[List[int]]) -> List[int]: + g = defaultdict(list) + indeg = [0] * numCourses + for a, b in prerequisites: + g[b].append(a) + indeg[a] += 1 + ans = [] + q = deque(i for i, x in enumerate(indeg) if x == 0) + while q: + i = q.popleft() + ans.append(i) + for j in g[i]: + indeg[j] -= 1 + if indeg[j] == 0: + q.append(j) + return ans if len(ans) == numCourses else [] +``` + +```python +#R 329 Longest Increasing Path in a Matrix/矩阵中的最长递增路径 (Medium) +class Solution: + def longestIncreasingPath(self, matrix: List[List[int]]) -> int: + m, n = len(matrix), len(matrix[0]) + flag = [[-1] * n for _ in range(m)] #存储从（i，j）出发的最长递归路径 + + def dfs(i, j): + if flag[i][j] != -1: # 记忆化搜索，避免重复的计算 + return flag[i][j] + else: + d = 1 + for (x, y) in [[-1, 0], [1, 0], [0, 1], [0, -1]]: + x, y = i + x, j + y + if 0 <= x < m and 0 <= y < n and matrix[x][y] > matrix[i][j]: + d = max(d, dfs(x, y) + 1) # 取四个邻接点的最长 + flag[i][j] = d + return d + + res = 0 + for i in range(m): # 遍历矩阵计算最长路径 + for j in range(n): + if flag[i][j] == -1: + res = max(res, dfs(i, j)) + return res +``` + +```python +#R 743 Network Delay Time/网络延迟时间 (Medium) +# 从某个节点 K 发出一个信号。需要多久才能使所有节点都收到信号 Dijkstra 算法 +class Solution: + def networkDelayTime(self, times: List[List[int]], n: int, k: int) -> int: + g = [[float('inf')] * n for _ in range(n)] + for x, y, time in times: + g[x - 1][y - 1] = time + dist = [float('inf')] * n + dist[k - 1] = 0 + used = [False] * n + for _ in range(n): + x = -1 + for y, u in enumerate(used): + if not u and (x == -1 or dist[y] < dist[x]): + x = y + used[x] = True + for y, time in enumerate(g[x]): + dist[y] = min(dist[y], dist[x] + time) + ans = max(dist) + return ans if ans < float('inf') else -1 +``` + +```python +#R 827 Making A Large Island/最大人工岛 (Hard) +# 最多 只能将一格 0 变成 1 +class Solution: + def largestIsland(self, grid: List[List[int]]) -> int: + n = len(grid) + def dfs(i: int, j: int) -> int: + size = 1 + grid[i][j] = len(area) + 2 # 记录 (i,j) 属于哪个岛 + for x, y in (i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1): + if 0 <= x < n and 0 <= y < n and grid[x][y] == 1: + size += dfs(x, y) + return size + # DFS 每个岛，统计各个岛的面积，记录到 area 列表中 + area = [] + for i, row in enumerate(grid): + for j, x in enumerate(row): + if x == 1: + area.append(dfs(i, j)) + # 加上这个特判，可以快很多 + if not area: # 没有岛 + return 1 + ans = 0 + for i, row in enumerate(grid): + for j, x in enumerate(row): + if x: continue + s = set() + for x, y in (i - 1, j), (i + 1, j), (i, j - 1), (i, j + 1): + if 0 <= x < n and 0 <= y < n and grid[x][y]: + s.add(grid[x][y]) # 记录上下左右格子所属岛屿编号 + ans = max(ans, sum(area[idx - 2] for idx in s) + 1) # 累加面积 + return ans if ans else n * n # 如果最后 ans 仍然为 0，说明所有格子都是 1，返回 n^2 +``` + + +```python +#R 0 Get Watched Videos by Your Friends/获取你好友已观看的视频 (Medium) +# watchedVideos[i] 和 friends[i] 分别表示 id = i 的人观看过的视频列表和他的好友列表。找出所有指定 level 的视频 +class Solution: + def watchedVideosByFriends(self, watchedVideos, friends, id, level): + n = len(friends) + used = [False] * n + q = collections.deque([id]) + used[id] = True + for _ in range(level): + span = len(q) + for i in range(span): + u = q.popleft() + for v in friends[u]: + if not used[v]: + q.append(v) + used[v] = True + freq = collections.Counter() + for _ in range(len(q)): + u = q.pop() + for watched in watchedVideos[u]: + freq[watched] += 1 + videos = list(freq.items()) + videos.sort(key=lambda x: (x[1], x[0])) + ans = [video[0] for video in videos] + return ans +``` + +```python +#R 1761 Minimum Degree of a Connected Trio in a Graph/一个图中连通三元组的最小度数 (困难) +# 连通三元组的度数 是所有满足此条件的边的数目：一个顶点在这个三元组内，而另一个顶点不在这个三元组内。返回所有连通三元组中度数的 最小值 +class Solution: + def minTrioDegree(self, n: int, edges: List[List[int]]) -> int: + g = [[False] * n for _ in range(n)] + deg = [0] * n + for u, v in edges: + u, v = u - 1, v - 1 + g[u][v] = g[v][u] = True + deg[u] += 1 + deg[v] += 1 + ans = inf + for i in range(n): + for j in range(i + 1, n): + if g[i][j]: + for k in range(j + 1, n): + if g[i][k] and g[j][k]: + ans = min(ans, deg[i] + deg[j] + deg[k] - 6) + return -1 if ans == inf else ans +``` + + +# 动态规划 + +简单 + + +```python +#R 背包问题 +def test_2_ei_bag_problem1(weight, value, bagweight): + # 二维数组 + dp = [[0] * (bagweight + 1) for _ in range(len(weight))] + # 初始化 + for j in range(weight[0], bagweight + 1): + dp[0][j] = value[0] + # weight数组的大小就是物品个数 + for i in range(1, len(weight)): # 遍历物品 + for j in range(bagweight + 1): # 遍历背包容量 + if j < weight[i]: + dp[i][j] = dp[i - 1][j] + else: + dp[i][j] = max(dp[i - 1][j], dp[i - 1][j - weight[i]] + value[i]) + return dp[len(weight) - 1][bagweight] +``` + + + +```python +#R 零钱兑换 计算可以凑成总金额所需的最少的硬币个数。 +class Solution: + def coinChange(self, coins: List[int], amount: int) -> int: + n = len(coins) + dp = [[amount+1] * (amount+1) for _ in range(n+1)] # 初始化为一个较大的值，如 +inf 或 amount+1 + # 合法的初始化 + dp[0][0] = 0 # 其他 dp[0][j]均不合法 + # 完全背包：优化后的状态转移 + for i in range(1, n+1): # 第一层循环：遍历硬币 + for j in range(amount+1): # 第二层循环：遍历背包 + if j < coins[i-1]: # 容量有限，无法选择第i种硬币 + dp[i][j] = dp[i-1][j] + else: # 可选择第i种硬币 + dp[i][j] = min( dp[i-1][j], dp[i][j-coins[i-1]] + 1 ) + ans = dp[n][amount] + return ans if ans != amount+1 else -1 +``` + +```python +#R 1. 买卖股票的最佳时机 +class Solution: + def maxProfit(self, prices: List[int]) -> int: + length = len(prices) + if len == 0: + return 0 + dp = [[0] * 2 for _ in range(length)] + dp[0][0] = -prices[0] + dp[0][1] = 0 + for i in range(1, length): + dp[i][0] = max(dp[i-1][0], -prices[i]) + dp[i][1] = max(dp[i-1][1], prices[i] + dp[i-1][0]) + return dp[-1][1] +``` + +```python +#R 122.买卖股票的最佳时机II +# 可以尽可能地完成更多的交易，不能同时参与多笔交易 +class Solution: + def maxProfit(self, prices: List[int]) -> int: + length = len(prices) + dp = [[0] * 2 for _ in range(length)] + dp[0][0] = -prices[0] + dp[0][1] = 0 + for i in range(1, length): + dp[i][0] = max(dp[i-1][0], dp[i-1][1] - prices[i]) #注意这里是和121. 买卖股票的最佳时机唯一不同的地方 + dp[i][1] = max(dp[i-1][1], dp[i-1][0] + prices[i]) + return dp[-1][1] +``` + +```python +#R 0 买卖股票的最佳时机含手续费 +class Solution: + def maxProfit(self, prices: List[int], fee: int) -> int: + n = len(prices) + dp = [[0] * 2 for _ in range(n)] + dp[0][0] = -prices[0] #持股票 + for i in range(1, n): + dp[i][0] = max(dp[i-1][0], dp[i-1][1] - prices[i]) + dp[i][1] = max(dp[i-1][1], dp[i-1][0] + prices[i] - fee) + return max(dp[-1][0], dp[-1][1]) +``` + +```python +#R 123.买卖股票的最佳时机III 最多可以完成 两笔 交易。 +# 0 没有操作 （其实我们也可以不设置这个状态）； 1 第一次持有股票； 2 第一次不持有股票 +class Solution: + def maxProfit(self, prices: List[int]) -> int: + if len(prices) == 0: + return 0 + dp = [[0] * 5 for _ in range(len(prices))] + dp[0][1] = -prices[0] + dp[0][3] = -prices[0] + for i in range(1, len(prices)): + dp[i][0] = dp[i-1][0] + dp[i][1] = max(dp[i-1][1], dp[i-1][0] - prices[i]) + dp[i][2] = max(dp[i-1][2], dp[i-1][1] + prices[i]) + dp[i][3] = max(dp[i-1][3], dp[i-1][2] - prices[i]) + dp[i][4] = max(dp[i-1][4], dp[i-1][3] + prices[i]) + return dp[-1][4] +``` + +```python +#R 188.买卖股票的最佳时机IV +最多可以完成 k 笔交易 +class Solution: + def maxProfit(self, k: int, prices: List[int]) -> int: + if len(prices) == 0: + return 0 + dp = [[0] * (2*k+1) for _ in range(len(prices))] + for j in range(1, 2*k, 2): + dp[0][j] = -prices[0] + for i in range(1, len(prices)): + for j in range(0, 2*k-1, 2): + dp[i][j+1] = max(dp[i-1][j+1], dp[i-1][j] - prices[i]) + dp[i][j+2] = max(dp[i-1][j+2], dp[i-1][j+1] + prices[i]) + return dp[-1][2*k] +``` + +```python +#R 1. 股票最佳买卖股票时机含冷冻期 +class Solution: + def maxProfit(self, prices: List[int]) -> int: + if not prices: + return 0 + n = len(prices) + # f[i][0]: 手上持有股票的最大收益 + # f[i][1]: 手上不持有股票，并且处于冷冻期中的累计最大收益 + # f[i][2]: 手上不持有股票，并且不在冷冻期中的累计最大收益 + f = [[-prices[0], 0, 0]] + [[0] * 3 for _ in range(n - 1)] + for i in range(1, n): + f[i][0] = max(f[i - 1][0], f[i - 1][2] - prices[i]) + f[i][1] = f[i - 1][0] + prices[i] + f[i][2] = max(f[i - 1][1], f[i - 1][2]) + return max(f[n - 1][1], f[n - 1][2]) +``` + + +子序列系列 + +```python +#R 0 不同的子序列 +# s 的子序列中 t 出现的个数 +class Solution: + def numDistinct(self, s: str, t: str) -> int: + dp = [[0] * (len(t)+1) for _ in range(len(s)+1)] + for i in range(len(s)): + dp[i][0] = 1 + for j in range(1, len(t)): + dp[0][j] = 0 + for i in range(1, len(s)+1): + for j in range(1, len(t)+1): + if s[i-1] == t[j-1]: + dp[i][j] = dp[i-1][j-1] + dp[i-1][j] + else: + dp[i][j] = dp[i-1][j] + return dp[-1][-1] +``` + +```python +#R 0 最长回文子序列 +class Solution: + def longestPalindromeSubseq(self, s: str) -> int: + dp = [[0] * len(s) for _ in range(len(s))] + for i in range(len(s)): + dp[i][i] = 1 + for i in range(len(s)-1, -1, -1): + for j in range(i+1, len(s)): + if s[i] == s[j]: + dp[i][j] = dp[i+1][j-1] + 2 + else: + dp[i][j] = max(dp[i+1][j], dp[i][j-1]) + return dp[0][-1] +``` + + + + +```python +#R 33.9 Wildcard Matching/通配符匹配 (Hard) +# '?' 可以匹配任何单个字符。'*' 可以匹配任意字符序列（包括空字符序列）。 +class Solution: + def isMatch(self, s: str, p: str) -> bool: + m, n = len(s), len(p) + f = [[False] * (n + 1) for _ in range(m + 1)] + # 初始条件：翻译自递归边界 + # 对应边界 i<0 and j < 0 return True + f[0][0] = True + # 对应边界 i<0 j>=0的处理 + for j in range(n): + f[0][j + 1] = p[j] == '*' and f[0][j] + # 对应边界 j<0 (f初始化已经赋值了False, 不用写) + for i in range(m): + for j in range(n): + if s[i] == p[j] or p[j] == '?': + f[i + 1][j + 1] = f[i][j] + elif p[j] == '*': + f[i + 1][j + 1] = f[i][j + 1] or f[i + 1][j] + return f[m][n] +``` + +```python +#R 91 Decode Ways/解码方法 (Medium) +# "12" 可以解码为 "AB"（1 2）或者 "L"（12） 时间复杂度O(n),空间复杂度O(n) +class Solution: + def numDecodings(self, s: str) -> int: + size = len(s) + #特判 + if size == 0: + return 0 + dp = [0]*(size+1) + dp[0] = 1 + for i in range(1,size+1): + t = int(s[i-1]) + if t>=1 and t<=9: + dp[i] += dp[i-1] #最后一个数字解密成一个字母 + if i >=2:#下面这种情况至少要有两个字符 + t = int(s[i-2])*10 + int(s[i-1]) + if t>=10 and t<=26: + dp[i] += dp[i-2]#最后两个数字解密成一个一个字母 + return dp[-1] +``` + +```python +#R 311.66 Burst Balloons/戳气球 (Medium) +# nums = [3,1,5,8] --> [3,5,8] --> [3,8] --> [8] --> [] +# 定义 f[i][j] 表示戳破区间 [i,j] 内的所有气球能得到的最多硬币数 +class Solution: + def maxCoins(self, nums: List[int]) -> int: + n = len(nums) + arr = [1] + nums + [1] + f = [[0] * (n + 2) for _ in range(n + 2)] + for i in range(n - 1, -1, -1): + for j in range(i + 2, n + 2): + for k in range(i + 1, j): + f[i][j] = max(f[i][j], f[i][k] + f[k][j] + arr[i] * arr[k] * arr[j]) + return f[0][-1] +``` + +```python +#R 377 Combination Sum IV/组合总和 Ⅳ (Medium) +# 与39题不同之处，顺序不同的序列被视作不同的组合。 +class Solution(object): + def combinationSum4(self, nums, target): + dp = [0] * (target + 1) + dp[0] = 1 + res = 0 + for i in range(target + 1): + for num in nums: + if i >= num: + dp[i] += dp[i - num] + return dp[target] +``` + +```python +#H 0 Arithmetic Slices II - Subsequence/等差数列划分 II - 子序列 (Hard) +# 所有等差子序列的数目 dp_ik=\sum_j^i-1 (dp_jk+1) 第一维用哈希表提高查询效率、节省空间 +class Solution: + def numberOfArithmeticSlices(self, nums: List[int]) -> int: + n,ans=len(nums),0 + dp=[defaultdict(int) for _ in range(n)] + for i,num in enumerate(nums): + for j in range(i): + k=num-nums[j] + dp[i][k]+=dp[j][k]+1 + ans+=dp[j][k] + return ans +``` + +```python +#H 0 Longest Arithmetic Subsequence/最长等差数列 (Medium) +# 以 a[i] 结尾的最长等差子序列公差及其长度，存到一个哈希表dp(i)={d:L} +class Solution: + def longestArithSeqLength(self, a: List[int]) -> int: + f = [{} for _ in range(len(a))] + for i, x in enumerate(a): + for j in range(i - 1, -1, -1): + d = x - a[j] # 公差 + if d not in f[i]: + f[i][d] = f[j].get(d, 1) + 1 + return max(max(d.values()) for d in f[1:]) +``` + +```python +#H 873 Length of Longest Fibonacci Subsequence/最长的斐波那契子序列的长度 (Medium) +# 这f[x][y]代表了以数字x和y结尾的最大斐波那契序列长度。f[x][y]=f[y−x][x]+1 由于数据范围很大，用哈希表嵌套哈希表实现。 +class Solution: + def lenLongestFibSubseq(self, A: List[int]) -> int: + + dp = {} + res = 0 + tempA = set(A) + for i in range(1,len(A)): + for j in range(i): + diff = A[i]-A[j] + if diff int: + if not s: + return 0 + dp = {} # 记录以某个字符结尾的最长连续子串长度 + max_len = 0 # 当前连续子串长度 + for i in range(len(s)): + if i > 0 and (ord(s[i]) - ord(s[i-1]) == 1 or (s[i-1] == 'z' and s[i] == 'a')): + max_len += 1 + else: + max_len = 1 + dp[s[i]] = max(dp.get(s[i], 0), max_len) + return sum(dp.values()) +``` + +```python +#R 688 Knight Probability in Chessboard/骑士在棋盘上的概率 (Medium) +# dp[step][i][j]= 1/8 sum_di,dj dp[step−1][i+di][j+dj] +class Solution: + def knightProbability(self, n: int, k: int, row: int, column: int) -> float: + dp = [[[0] * n for _ in range(n)] for _ in range(k + 1)] + for step in range(k + 1): + for i in range(n): + for j in range(n): + if step == 0: + dp[step][i][j] = 1 + else: + for di, dj in ((-2, -1), (-2, 1), (2, -1), (2, 1), (-1, -2), (-1, 2), (1, -2), (1, 2)): + ni, nj = i + di, j + dj + if 0 <= ni < n and 0 <= nj < n: + dp[step][i][j] += dp[step - 1][ni][nj] / 8 + return dp[k][row][column] +``` + +```python +#R 877 Stone Game/石子游戏 (Medium) +class Solution: + def stoneGame(self, piles: List[int]) -> bool: + N = len(piles) + f = [[0]*(N+1) for _ in range(N+1)] # 防止出界 + for l in range(N): + for i in range(N-l): + j = i+l + f[i][j] = max(piles[i]-f[i+1][j], + piles[j]-f[i][j-1]) + return f[0][N-1]>0 +``` + +```python +#R 1000 Minimum Cost to Merge Stones/合并石头的最低成本 (Hard) +# 每次移动需要将连续的k堆石头合并为一堆，成本为这k堆中石头的总数。 +# 定义 f[i][j][k] 表示将 [i,j] 合并成k堆的最小成本 dp[i][j][k] = min(dp[i][m][1] + dp[m+1][j][k-1]) i≤m int: + n = len(stones) + if (n - 1) % (K-1): return -1 + def get(i, j): + return sums[j+1] - sums[i] + dp = [[[float("inf")] * (K+1) for _ in range(n)] for _ in range(n)] + sums = [0] * (1+n) + for i in range(1, n+1): sums[i] = sums[i-1] + stones[i-1] + for i in range(n): dp[i][i][1] = 0 + for l in range(2, n+1): + for i in range(n - l + 1): + j = i + l - 1 + for k in range(2, K+1): + for m in range(i, j): + dp[i][j][k] = min(dp[i][j][k], dp[i][m][1] + dp[m+1][j][k-1]) + dp[i][j][1] = dp[i][j][K] + get(i, j) #合并成一堆特殊情况不可以从合并成0获得 + return dp[0][n-1][1] +``` + + +```python +#R 1395 Count Number of Teams/统计作战单位数 (中等) +# 3个士兵组成一个作战单位单调的作战单位的方案数。 +# dp[i]记录的是第i个数之前比其值小的数的个数 +class Solution: + def numTeams(self, rating: List[int]) -> int: + def func(nums): + dp = [0] * len_ + res = 0 + for i in range(1, len_): + idx = i - 1 + while idx >= 0: + if nums[i] > nums[idx]: + dp[i] += 1 + if dp[idx] > 0: + res += dp[idx] + idx -= 1 + return res + len_ = len(rating) + return func(rating[::-1]) + func(rating) +``` + + +# 其他 + +