【LeetCode】Remove Nth Node From End of List

Given a linked list, remove the nth node from the end of list and return its head.

For 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.
Try to do this in one pass.

思路：双指针，一个指针先向前走n步，一个指针在跟上一起向前走到NULL，后一个指针之后的即为删除节点。

/** * 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) {        if(NULL == head)return head;        ListNode *pre=NULL;        ListNode *flag=head;        ListNode *ret=head;        for(int i=0;i<n-1;i++){            flag=flag->next;        }                while(NULL != flag->next){            pre=ret;            flag=flag->next;            ret=ret->next;        }        if(NULL == pre){            head=head->next;            delete ret;        }else{            pre->next=ret->next;            delete ret;        }        return head;    }};