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。

1 /** 2  * Definition for singly-linked list. 3  * struct ListNode { 4  *     int val; 5  *     ListNode *next; 6  *     ListNode(int x) : val(x), next(NULL) {} 7  * }; 8  */ 9 class Solution {10 public:11     ListNode *removeNthFromEnd(ListNode *head, int n) {12         // Start typing your C/C++ solution below13         // DO NOT write int main() function14         if (head == NULL)15             return NULL;16             17         ListNode *pPre = NULL;18         ListNode *p = head;19         ListNode *q = head;20         for(int i = 0; i < n - 1; i++)21             q = q->next;22             23         while(q->next)24         {25             pPre = p;26             p = p->next;27             q = q->next;28         }29         30         if (pPre == NULL)31         {32             head = p->next;33             delete p;34         }35         else36         {37             pPre->next = p->next;38             delete p;39         }40         41         return head;42     }43 };