随笔分类 - leetcode
摘要：题目描述： 方法一：二分 时间复杂度应该是O(n2) class Solution: def countSmaller(self, nums: List[int]) -> List[int]: num_length = len(nums) if not nums: return [] res = [
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摘要：题目描述： 提交： class Solution: def reformatDate(self, date: str) -> str: l = date.split(" ") d = l[0][:-2] ml = ["Jan", "Feb", "Mar", "Apr", "May", "Jun",
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摘要：题目描述： 方法一：动态规划 O(n) O(n) class Solution: def maxProfit(self, prices: List[int]) -> int: if len(prices) < 2: return 0 n = len(prices) sell = [0] * n bu
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摘要：题目描述： 方法一：动态规划 O(n2) ->O(mn) m为字典中单词最大长度 class Solution: def respace(self, dictionary: List[str], sentence: str) -> int: d = {}.fromkeys(dictionary) n
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摘要：题目描述： 方法一：纯volatile实现 class FizzBuzz { private int n; private volatile int f = 0; private volatile int b = 0; private volatile int fb = 0; private vol
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摘要：题目描述： 提交：O(n) class Solution: def divingBoard(self, shorter: int, longer: int, k: int) -> List[int]: if k == 0: return [] res = [] for i in range(k,-1
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摘要：题目描述： 方法一：动态规划 O(n) O(n) class Solution: def longestValidParentheses(self, s: str) -> int: n = len(s) if n==0:return 0 dp = [0]*n for i in range(len(s
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摘要：题目描述： 方法一：动态规划 O(mn) O(mn) class Solution: def findLength(self, A: List[int], B: List[int]) -> int: n, m = len(A), len(B) dp = [[0] * (m + 1) for _ in
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摘要：题目描述： 方法：动态规划+状态压缩 class Solution: def minNumberOfSemesters(self, n: int, dependencies: List[List[int]], k: int) -> int: dep = {} # 记录依赖于某节点的节点列表 for
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摘要：题目描述： 方法一：原地哈希 class Solution: def firstMissingPositive(self, nums: List[int]) -> int: n = len(nums) for i in range(n): if nums[i] <= 0: nums[i] = n +
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摘要：# Definition for singly-linked list. # class ListNode: # def __init__(self, x): # self.val = x # self.next = None class Solution: def removeDuplicateN
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摘要：题目描述： 提交： class Solution: from typing import List def avoidFlood(self, rains: List[int]) -> List[int]: import heapq heap = [] res = [-1 if i != 0 else
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摘要：题目描述： 第一次提交： 有错 未找出问题 留坑 class Solution: def getFolderNames(self, names: List[str]) -> List[str]: res = [] dic = collections.Counter() for name in nam
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摘要：方法一： class Solution: def patternMatching(self, pattern: str, value: str) -> bool: count_a = sum(1 for ch in pattern if ch == 'a') count_b = len(patter
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摘要：方法：递归 O(n) O(n) class Solution: def __init__(self): self.maxSum = float("-inf") def maxPathSum(self, root: TreeNode) -> int: def maxGain(node): if not
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摘要：提交： class Solution: def findLeastNumOfUniqueInts(self, arr: List[int], k: int) -> int: count = collections.Counter(arr) order_dic = sorted(count.items
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摘要：方法：倍增法dp ACM经典 class TreeAncestor: def __init__(self, n: int, parent: List[int]): self.cols = 20 # log(50000) < 20 self.dp = [[-1] * self.cols for _ i
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摘要：方法：迭代+栈 O(N) class Solution { public TreeNode recoverFromPreorder(String S) { Deque<TreeNode> path = new LinkedList<>(); int pos = 0; while(pos < S.le
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摘要：解：O(N) class Solution: def maxScoreSightseeingPair(self, A: List[int]) -> int: left, res = A[0], -1 for j in range(1, len(A)): res = max(res, left + A
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摘要：题目描述： 提交：二分 O(nlogk) class Solution: def minDays(self, bloomDay: List[int], m: int, k: int) -> int: minday = min(bloomDay) maxday = max(bloomDay) def
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