80 KiB
80 KiB
#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
#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
#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
#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
#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
#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
#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
#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
#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
#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)
#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
#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
#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
#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])
#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
#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
#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
#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
#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)
#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()
#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
#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。
#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)
#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
#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
#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]
#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
#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
#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]
#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
#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
#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
#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
#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
#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)
#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
#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
#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]
#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
#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
#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
#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)
#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
#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)
#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
#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
#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()]
#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)
#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
#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
#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
#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
#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
#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
#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()
#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
#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
#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
#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
#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
#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
#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
#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
#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
#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
#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
#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 []
#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
#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
#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
#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
#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
#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]
#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
#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]
#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]
#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])
#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]
#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]
#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
#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]
#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]
#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]
#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]
#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]
#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]
#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
#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:])
#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 <A[j] and diff in tempA:
dp[(A[j],A[i])] = dp.get((diff,A[j]),2)+1
res = max(res,dp[(A[j],A[i])])
return res
#R 467 Unique Substrings in Wraparound String/环绕字符串中唯一的子字符串 (Medium) 字符串 s = "zab" 有六个子字符串 ("z", "a", "b", "za", "ab", and "zab") 在 base 中
# dp[i] 表示以 s[i] 结尾的最长连续有效子串的长度。
class Solution:
def findSubstringInWraproundString(self, s: str) -> 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())
#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]
#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
#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<j
def mergeStones(self, stones: List[int], K: int) -> 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]
#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)