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一、leetcode-19

删除链表的倒数第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。

输入：head = [1,2,3,4,5], n = 2

输出：[1,2,3,5]

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二、题解

1.引库

 #include <iostream>#include <cstdio>#include <cstdlib>#include <queue>#include <stack>#include <algorithm>#include <string>#include <map>#include <set>#include <vector>using namespace std;

2.代码

/*** Definition for singly-linked list.* struct ListNode {*     int val;*     ListNode *next;*     ListNode() : val(0), next(nullptr) {}*     ListNode(int x) : val(x), next(nullptr) {}*     ListNode(int x, ListNode *next) : val(x), next(next) {}* };*/
class Solution {
public:ListNode* removeNthFromEnd(ListNode* head, int n) {ListNode *newhead=new ListNode(0),*fast=newhead,*slow=newhead;newhead->next=head;for(int i=0;i<=n;i++) fast=fast->next;while(fast!=NULL) fast=fast->next,slow=slow->next;ListNode *tmp=slow->next;slow->next=slow->next->next;delete tmp;return newhead->next;}
};