力扣题目链接

题目链接/文章讲解/视频讲解：https://programmercarl.com/0019.%E5%88%A0%E9%99%A4%E9%93%BE%E8%A1%A8%E7%9A%84%E5%80%92%E6%95%B0%E7%AC%ACN%E4%B8%AA%E8%8A%82%E7%82%B9.html

1.说真的不看代码随想录这个题我是真的想不到还有这么妙的思路呢，先让fast指针先到n+1的位置，然后同时移动fast slow指针就可以让slow指针指向要删除的节点的前一个节点，这样就可以将这个节点巧妙的删除呢。

2.这个题给我的收获远不止这个思路，而且还有内存回收的思想，还有检查是否有空指针异常的问题，真的让我受益匪浅。

CPP

/**
 * 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 *dummyHead = new ListNode(0);
        dummyHead->next = head;
        ListNode *slow = dummyHead;
        ListNode *fast = dummyHead;
        
        while(n-- && fast != NULL){
            fast = fast->next;
        }
        fast = fast->next;
        while(fast != NULL){
            fast = fast->next;
            slow = slow->next;
        }
        
        ListNode *temp = new ListNode(0);
        temp = slow->next;
        slow->next = slow->next->next;
        delete temp;
        ListNode *res = new ListNode(0);
        res = dummyHead->next;
        delete dummyHead;
        return res;
    }
};