Leetcode-19 Remove Nth Node From End of List

发表于 | 分类于 | 次

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.

Note:

Given n will always be valid.

Follow up:

Could you do this in one pass?

Example 1

1
2
3
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.

解题思路

双指针 p 和 q。两指针相距 n，移动两个指针直至 q 到达末尾。此时 p 指向待删除节点的前面。当删除第一个节点时，需要增加哑节点。

复杂度分析

  • 时间复杂度：$O(n)$
  • 空间复杂度：$O(1)$

代码

1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
/**
 * 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 = new ListNode(0), *q = p;
		p->next = head;
		head=p;
		while (q->next) {
			if (n-- <= 0) 
				p=p->next;
			q=q->next;
		}
		p->next = p->next->next;
		return head->next;
	}
};