cmaster1219的博客

我的解法：class Solution: def reverseBetween(self, head: ListNode, m: int, n: int) -> ListNode: count = 0 dummy = ListNode() dummy.next = head p = dummy inv_nodes = [] while count <= n: if .

我的解法：class Solution: def spiralOrder(self, matrix: List[List[int]]) -> List[int]: if not matrix: return [] if len(matrix)==1: return matrix[0] m,n = len(matrix),len(matrix[0]) directions = [(.

我的解法：class Solution: def removeNthFromEnd(self, head: ListNode, n: int) -> ListNode: dummyhead = ListNode(None) dummyhead.next = head cur = dummyhead for i in range(n+1): cur = cur.next tail = du.

我的解法：class Solution: def trailingZeroes(self, n: int) -> int: fives = 0 while n > 0: n = n//5 fives += n return fives末尾的一个零表示阶乘中至少有一对2和5的因子，由于2因子的数量远多于5因子，因此只需考虑阶乘的所有乘数中一共有多少个5因子。需要注意的是有的乘数会包含多个.

我的解法：# Definition for a binary tree node.# class TreeNode:# def __init__(self, x):# self.val = x# self.left = None# self.right = Noneclass Solution: def isSameTree(self, p: TreeNode, q: TreeNode) -> bool: .

我的解法：class Solution: def countAndSay(self, n: int) -> str: if n == 1: return "1" i = 1 obj = self.countAndSay(n-1) val = obj[0] count = 1 res = '' while i < len(obj): .

我的解法：def twoSum(self, nums: List[int], target: int) -> List[int]: for i, num in enumerate(nums): for j in range(i+1,len(nums)): if nums[j] == target-num: return [i,j]两层循环，时间复杂度O(n2)大佬解法：def twoSum(self, nums.