【LeetCode】LeetCode——第19题：Remove Nth Node From End of List

19. Remove Nth Node From End of List

    My Submissions

Question Editorial Solution

Total Accepted: 106851  Total Submissions: 362053  Difficulty: Easy

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.

题目的大概意思是：给定一个单向链表，删除其倒数第n个结点。

这道题难度等级为：简单

思路：遍历链表的长度，然后再次遍历找到倒数第n个结点，将其删除。

注意：输入的链表是没有头结点的。

代码如下：

/**
 * 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* p = head;
		ListNode* q = head;
        int len = 0, cnt = 0;
		while (p != NULL){	//表长
			p = p->next;
			++len;
		}
		if (len == n){return head->next;}
		while(cnt < len - n - 1){
			q = q->next;
			++cnt;
		}
		(n == 1) ? (q->next = NULL) : (q->next = q->next->next);
		return head;
    }
};

提交代码，AC时间为Runtime:4ms。