随笔分类 - leetcode
摘要：题目描述： 方法一： class Solution: def maxProfit(self, k: int, prices: List[int]) -> int: if len(prices) <= 1: return 0 if (k < len(prices) // 2) : dp = [[-pr
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摘要：题目描述： 方法一： class Solution: def maxProfit(self, prices: List[int]) -> int: dp_i1_0 = 0 dp_i1_1 = float('-inf') dp_i2_0 = 0 dp_i2_1 = float('-inf') for
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摘要：题目描述： 方法一： class Solution: def maxProfit(self, prices: List[int]) -> int: profit = 0 for i in range(1,len(prices)): tem = prices[i] - prices[i-1] if t
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摘要：题目描述; 方法一：O(n） O(1） class Solution: def maxProfit(self, prices: List[int]) -> int: minprices = float('inf') maxprofit = 0 for i in prices: if i<minpri
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摘要：题目描述： 第一次提交：动态规划 class Solution: def minimumTotal(self, triangle: List[List[int]]) -> int: n = len(triangle) for i in range(n-2,-1,-1): for j in range
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摘要：题目描述： 第一次提交： class Solution: def getRow(self, rowIndex: int) -> List[int]: k = rowIndex pre = [1] + [0] * k res = [1] + [0] * k for i in range(1,k+1):
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摘要：题目描述： 输入：{"$id":"1","left":{"$id":"2","left":{"$id":"3","left":null,"next":null,"right":null,"val":4},"next":null,"right":{"$id":"4","l
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摘要：题目描述： 方法一：迭代 class Solution: def flatten(self, root: TreeNode) -> None: """ Do not return anything, modify root in-place instead. """ cur = root while
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摘要：恢复内容开始 题目描述： 方法一：O(n) O(n) class Solution: def sortedListToBST(self, head: ListNode) -> TreeNode: nums = [] while head: nums.append(head.val) head = h
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摘要：题目描述： 方法一：O(n) O(n) class Solution: def buildTree(self, inorder: List[int], postorder: List[int]) -> TreeNode: assert len(inorder)==len(postorder) if
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摘要：题目描述： 方法一： # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class
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摘要：题目描述： 方法一： # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None class
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摘要：题目描述： 方法一：递归O(n) O(n) # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right =
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摘要：题目描述： 方法一：递归 # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None cla
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摘要：题目描述： 方法一：动态规划 O(n^2) O(n) class Solution: def numTrees(self, n: int) -> int: dp = [0]*(n+1) dp[0],dp[1] = 1,1 for i in range(2,n+1): for j in range(1
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摘要：题目描述： 方法一：递归 class Solution: def postorderTraversal(self, root: TreeNode) -> List[int]: res = [] def helper(root): if not root: return helper(root.lef
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摘要：恢复内容开始 题目描述: 方法一：递归 class Solution: def preorderTraversal(self, root: TreeNode) -> List[int]: res = [] def helper(root): if not root: return res.appen
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摘要：题目描述： 方法一：迭代 class Solution: def inorderTraversal(self, root: TreeNode) -> List[int]: res,stack = [],[] while True: while root: stack.append(root) roo
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摘要：题目描述： 方法一：暴力法 class Solution: def restoreIpAddresses(self, s: str) -> List[str]: n = len(s) res = [] # 判读是否满足ip的条件 def helper(tmp): if not tmp or (tmp
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摘要：题目描述： 方法一: class Solution: def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode: dummy = ListNode(0) dummy.next = head pre = dummy for
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