leetcode之Remove Nth Node From End of List

题目：

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.

解答：

为了防止只有一个元素，又要删除的情况，需要加一个phead，便于操作

寻找倒数第n个元素的方法：直接看代码啦！

/**
 * 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) {
        ListNode *tmp1 = head;
        ListNode *tmp2 = new ListNode(0);
        ListNode *phead = tmp2;
        phead->next = head;
        while(n--)
            tmp1 = tmp1->next;
        while(tmp1 != NULL)
        {
            tmp2 = tmp2->next;
            tmp1 = tmp1->next;
        }
        ListNode* t = tmp2->next;
        tmp2->next = tmp2->next->next;
        delete t;
        return phead->next;
    }
};