力扣 简单 876.链表的中间结点
文章目录
- 题目介绍
- 题解
题目介绍
题解
法一：
class Solution {public ListNode middleNode(ListNode head) {ListNode cur = head;int n = 0;while (cur != null) {n++;cur = cur.next;}ListNode curr = head;for (int i = 0; i < n / 2; i++) {curr = curr.next;}return curr;}
}
法二：快慢指针
class Solution {public ListNode middleNode(ListNode head) {ListNode slow = head;ListNode fast = head;while(true){if(fast == null || fast.next == null){break;}slow = slow.next;fast = fast.next.next;}return slow;}
}