每日算法之十八： 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.

依然是双指针思想，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。

<span style="font-size:18px;">/** * 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) {        // Start typing your C/C++ solution below        // DO NOT write int main() function        if (head == NULL)            return NULL;                    ListNode *pPre = NULL;        ListNode *p = head;        ListNode *q = head;        for(int i = 0; i < n - 1; i++)            q = q->next;                    while(q->next)        {            pPre = p;            p = p->next;            q = q->next;        }                if (pPre == NULL)        {            head = p->next;            delete p;        }        else        {            pPre->next = p->next;            delete p;        }                return head;    }};</span>