删除链表中倒数第n个节点
给你一个链表,删除链表的倒数第 n 个结点,并且返回链表的头结点。
用例
输入:head = [1,2,3,4,5], n = 2
输出:[1,2,3,5]
输入:head = [1], n = 1
输出:[]
输入:head = [1,2], n = 1
输出:[1]
提示
- 链表中结点的数目为 sz
- 1 <= sz <= 30
- 0 <= Node.val <= 100
- 1 <= n <= sz
示例代码
解法1:普通解法
var removeNthFromEnd = function(head, n) {let len=0;let rootNode=head;while(head){len++;head=head.next;}if(len==1) return null;len=len-n+1;//截取的目标位置let i=1;head=rootNode;while(head){console.log(i,head.val)if(i==len-1){//删除下一个节点head.next=head.next.next;break;}if(i==len){//删除当前rootNode=head.next;break;}i++;head=head.next;}return rootNode;
};
解法2:辅助栈
/*** Definition for singly-linked list.* function ListNode(val, next) {* this.val = (val===undefined ? 0 : val)* this.next = (next===undefined ? null : next)* }*/
/*** @param {ListNode} head* @param {number} n* @return {ListNode}*/
var removeNthFromEnd = function(head, n) {let stack=[]let root=head;//入栈while(head){stack.push(head)head=head.next;}let next=null;for(let i=1;i<n;i++){next=stack.pop();}let del=stack.pop();//找到目标值let prev=stack.pop(); if(prev){prev.next=next;}else{root=next;}return root;
};
解法3: 双指针
/*** @param {ListNode} head* @param {number} n* @return {ListNode}*/
var removeNthFromEnd = function(head, n) {let dummy=new ListNode(0,head);let slow=dummy;let first=head;for(let i=0;i<n;i++){first=first.next;}while(first){first=first.next;slow=slow.next;}slow.next=slow.next.next;return dummy.next;
};
执行情况:双指针
Tip
越简单的题,使用越多的解法才能让自己对代码的理解得以增强哦