链表

https://leetcode.cn/problems/remove-nth-node-from-end-of-list/

使用快慢指针，让nxt指针先向后移动N次

这样nxt指针和pre指针之间的间隔则为N

然后，nxt和pre同时向后移动，当nxt移动到链表的尾部，pre依然和它间隔为N，那么此时的pre就是链表的倒数第N个节点了

/**
 * 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* pre = head;
        ListNode* nxt = head;
        for(int i=0;i<n;i++) {
            nxt = nxt->next;
        }
        if(nxt == nullptr) {
            head = head->next;
            return head;
        }

        while(nxt!=nullptr) {
            nxt = nxt->next;
            if(nxt == nullptr) {
                cout<<pre->val<<endl;
                pre->next = pre->next->next;
                return head;                
            }
            pre = pre->next;
        }  
        return head;
    }
};