写代码的高酱

1. 方法一 （1）算法思想：遍历一遍整个链表，求得链表的长度len后，再从头遍历链表len/2遍，即可到达链表的中点处。 （2）代码： public ListNode findMid(ListNode head) { int len = 0; ListNode curr = head; while (curr != null) { len++; curr = curr.next; } curr = head; for (int i = 0; i < len / 2; i++)

Problem Solution # 法一：差值法 class MinStack: def __init__(self): """ initialize your data structure here. """ self.min_v = float('inf') self.stack = [] ...

Problem Solution class Solution: def singleNumber(self, nums: List[int]) -> int: single_num = 0 for num in nums: single_num = single_num ^ num return single...

Problem Solution class Solution: def maxProfit(self, prices: List[int]) -> int: max_profit = 0 for i in range(len(prices) - 1): if prices[i+1] - prices[i] > 0: ...

Problem Solution class Solution: def maxProfit(self, prices: List[int]) -> int: min_price = float('inf') max_profit = 0 for price in prices: # 在最小点买入，在最高点卖出...

Problem Solution # Definition for a binary tree node. # class TreeNode: # def __init__(self, x): # self.val = x # self.left = None # self.right = None # 法一：递归方式 class Solu...

Problem Solution # 从后向前填充 class Solution: def merge(self, nums1: List[int], m: int, nums2: List[int], n: int) -> None: """ Do not return anything, modify nums1 in-place instead...

Problem Solution # 法一：优化前缀和 class Solution: def maxSubArray(self, nums: List[int]) -> int: n = len(nums) max_sum = nums[0] min_sum = sum = 0 sum = 0 for...

Problem Solution class Solution: def removeDuplicates(self, nums: List[int]) -> int: if nums: slow = 0 for fast in range(1, len(nums)): if nums[...

Problem Solution class Solution: def isValid(self, s: str) -> bool: stack = [] map = { ')': '(', '}': '{', ']': '[' } if len...