Remove Nth Node From End of List 从链表中删除倒数第N的节点

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.

采用双指针思想，两个指针相隔n-1，每次两个指针向后一步，当后面一个指针没有后继了，前面一个指针就是要删除的节点。

/** * 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 (head == NULL)            return NULL;                ListNode *preNode = NULL;        ListNode *pNode = head;        ListNode *qNode = head;                for(int i = 0;i<n-1;i++){            qNode = qNode->next;        }                while(qNode->next){            preNode = pNode;            pNode = pNode->next;            qNode = qNode->next;        }                if(preNode == NULL){            head = pNode->next;            delete pNode;        }else{            preNode->next = pNode->next;            delete pNode;        }                return head;    }};