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：

/**
 * 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;
    }
};