19. Remove Nth Node From End of List 解题记录

题目描述：

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.

解题思路：

我的解题思路是用递归函数，从最后开始返回节点的位置，然后删除。直接说不太好理解，看代码会清楚一点。

代码：

/**
 * Definition for singly-linked list.
 * struct ListNode {
 *     int val;
 *     ListNode *next;
 *     ListNode(int x) : val(x), next(NULL) {}
 * };
 */
class Solution {
public:
    int removeListNode(ListNode* cur, int n){
        if(!cur)
        //链表末尾返回数值
            return -1;
        int num = removeListNode(cur->next, n)+1;
        //当前节点的计数，最后一个节点是0
        if(num == n)
            cur->next = cur->next->next;
        return num;
    }
    ListNode* removeNthFromEnd(ListNode* head, int n) {
        ListNode* pre = new ListNode(0);
        pre->next = head;
        int sum = removeListNode(pre, n);//新建一个节点，next为head，防止删除节点在第一个的情况
        return pre->next;
    }
};