leetcode:Remove Nth Node From End of List

Given a linked list, remove the n^th 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.

分析：题意为给一个链表，删除链表末端第n个节点然后返回头节点。

思路：利用双指针思想，两个指针之间间隔n-1，每个指针相后走一步，直到后面一个指针没有后继元素，此时前一个指针就是所要删除的节点。

代码：

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/**  * 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 *p,*q,*temp;         p=head;         q=head;         temp=NULL;         for(int i=0; i<n-1; i++){             q=q->next;         }                   while(q->next){             temp=p;             q=q->next;             p=p->next;         }                   if(temp==NULL){   //这种情况下，即n=1，删除的是头节点             head=p->next;             delete p;         }         else{             temp->next=p->next;             delete p;         }                   return head;     } };