题目

 

代码

/** * Definition for singly-linked list. * struct ListNode { *     int val; *     ListNode *next; *     ListNode(int x) : val(x), next(NULL) {} * }; */class Solution {public:    ListNode* removeNthFromEnd(ListNode* head, int n) {      ListNode* slow = head,*fast=head;               for(int i=0;i
    
     next;        }        //如果删除的是第一个元素        if(fast==NULL)        {            ListNode* tem = head;            head = head->next;            delete tem;            return head;        }                while(fast->next)        {            fast = fast->next;            slow = slow->next;        }        //最后停止的时候，慢的指针指向的是要删除的点之前那个元素        ListNode* tem = slow->next;        slow->next = slow->next->next;        delete tem;        return head;            }};
    

思路

先用一个快指针前进n步，当快指针的下一个是nullptr时，慢指针的下一个则是要删除的元素