题目描述

Given a linked list, remove the n-th node from the end of list and return its head.
Example:
Given linked list: 1->2->3->4->5, and n = 2.

After removing the second node from the end, the linked list becomes 1->2->3->5.
Note:
Given n will always be valid.
Follow up:
Could you do this in one pass?

样例

1->2->3->4->5 
n = 2
输出  1->2->3->5

C++ 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 *dummy = new ListNode(-1);  // 首先新建一个链表的结点
        dummy->next = head;                     //令这个结点指向head
        ListNode *first = dummy, *second = dummy;   //同时设置双指针指向该节点
        for (int i = 0; i < n; i ++ )   //将first指针指向第n个数 second 指针不动,还是在最前面,此时first  //和second这个节点相差n
        {
            first = first->next;
        }
        //当first结点指向最后的空d结点时second结点刚好指向倒数第n个结点
            while (first -> next)
        {
            first = first->next;
            second = second->next;
        }
        //直接删掉这个倒数第n个结点
        second->next = second->next->next;
        return dummy->next;
    }
};