【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;
}
};