Remove all elements from a linked list of integers that have valueval. Example: Input:1->2->6->3->4->5->6,val= 6Output:1->2->3->4->5 说到删除,首先想到定义两个指针,分别指向要被删除的结点和该结点的前驱结点。这里还需要考虑头结点是需要删除结点的特殊情况。 /** * Definition for s...
leetcode---Remove Linked List Elements Remove all elements from a linked list of integers that have valueval. Example Given:1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6,val= 6 Return:1 --> 2 --> 3 --> 4 --> 5 Credits: Special thanks to@mithmattfor adding this problem a...
Can you solve this real interview question? Remove Linked List Elements - Given the head of a linked list and an integer val, remove all the nodes of the linked list that has Node.val == val, and return the new head. Example 1: [https://assets.leetc
代码一 1// 203. Remove Linked List Elements 2// https://leetcode.com/problems/remove-linked-list-elements/description/ 3// 使用虚拟头结点 4// 时间复杂度: O(n) 5// 空间复杂度: O(1) 6class Solution { 7public: 8 ListNode* removeElements(ListNode* head, int val) { 910 // 创建虚拟...
Remove all elements from a linked list of integers that have value val. Example : AI检测代码解析 Input:1->2->6->3->4->5->6,val=6Output:1->2->3->4->5 1. 2. 题目大意 删除链表中所有指定值的结点。 解题思路 按照题意做即可。
https://leetcode-cn.com/problems/remove-linked-list-elements/description/ 删除链表里指定的值。这道题两个关键点: 1、遍历链表的时候需要保存当前节点的前一个节点的指针,在链表里只要涉及到删除元素的时候,都需要保存前一个节点的指针; 2、虚拟头节点的“秒”用,不需要对第一个元素做特殊处理,避免边界条件...
Remove all elements from a linked list of integers that have value val. Example Given: 1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6, val = 6 Return: 1 --> 2 --> 3 --> 4 --> 5 一刷 题解:很简单,直接上代码吧。
203. Remove Linked List Elements Remove all elements from a linked list of integers that have valueval. Example Given:1 --> 2 --> 6 --> 3 --> 4 --> 5 --> 6,val= 6 Return:1 --> 2 --> 3 --> 4 --> 5 题目大意: ...
【LeetCode203】删除链表中的结点(Remove Linked List Elements) Makaay SJTU 撸铁 人工智能 程序猿 神经质患者 3 人赞同了该文章 题目描述:删除链表中等于给定值 val 的所有节点。 示例: 输入: 1->2->6->3->4->5->6, val = 6 输出: 1->2->3->4->5思路解析(第一次):...
func removeElements(head *ListNode, val int) *ListNode { if head == nil { return head } var p1 ListNode p1.Next = head p2 := &p1 for head != nil { if head.Val == val { p2.Next, head = head.Next, head.Next } else { p2, head = head, head.Next } } return p1.Next ...