作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 设置当前节点的值为下一个 日期 [LeetCode] 题目地址:https://leetcode.com/problems/delete-node-in-a-linked-list/ Total Accepted: 78258 Total Submissions: 179086 Difficulty: Easy 题目描述 Write a functio…

题目: Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked list should become 1 -> 2 -…

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#-*- coding: UTF-8 -*- # Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):       def deleteNode(self, node):        if node.next==None or nod…

作者: 负雪明烛 id: fuxuemingzhu 个人博客:http://fuxuemingzhu.cn/ 目录 题目描述 题目大意 解题方法 迭代 日期 题目地址:https://leetcode.com/problems/delete-node-in-a-bst/description/ 题目描述 Given a root node reference of a BST and a key, delete the node with the given key in the BST. Re…

Given a root node reference of a BST and a key, delete the node with the given key in the BST. Return the root node reference (possibly updated) of the BST. Basically, the deletion can be divided into two stages: Search for a node to remove. If the n…

------------------------------------------------ 因为不知道前序是谁,所以只好采用类似于数组实现的列表移动值, 又因为如果当前是最后一个元素了但是已经没办法修改前序了所以必须在倒数第二个就修改,所以应该提前进行判断 AC代码: /** * Definition for singly-linked list. * public class ListNode { * int val; * ListNode next; * ListNode(int x)…

237 - Delete Node in a Linked List Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the linked…

237. Delete Node in a Linked List[easy] Write a function to delete a node (except the tail) in a singly linked list, given only access to that node. Supposed the linked list is 1 -> 2 -> 3 -> 4 and you are given the third node with value 3, the l…

283. Move Zeroes var moveZeroes = function(nums) { var num1=0,num2=1; while(num1!=num2){ nums.forEach(function(x,y){ if(x===0){ nums.splice(y,1); nums.push(0); } num1 = nums ; }); nums.forEach(function(x,y){ if(x===0){ nums.splice(y,1); nums.push(0);…