```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