链表_删除链表的倒数第N个节点

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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;
    }
};