19. 删除链表的倒数第N个节点

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<n;i++)
        {
            fast = fast->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