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：这道题从尾结点开始删除给定索引的值，然后再输出删除后的结点。我们可以使用两个指针pPost,pPre来解这道题，而且还要保存后面指针的前一结点为删除指定结点做准备。想让pPre先走n-1步，然后循环pPre和pPost共同起步行走，每一次之前都要把pPost保存下来，直到pPre->next为空循环结束。然后判断pTemp。

/**
 * 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(head==NULL)
            return NULL;
        ListNode *pTemp=NULL;
        ListNode *pPre=head;
        ListNode *pPost=head;
        for(int i=0;i<n-1;i++)
            pPre=pPre->next;
        while(pPre->next!=NULL)
        {
            pTemp=pPost;
            pPre=pPre->next;
            pPost=pPost->next;
        }
        if(pTemp==NULL)
        {
            head=head->next;
        }
        else
        {
            pTemp->next=pPost->next;
        }
        return head;
    }
};