LeetCode(876) 链表的中间节点

Given a non-empty, singly linked list with head node head, return a middle node of linked list.

If there are two middle nodes, return the second middle node.

给定一个单链表，返回链表的中间节点，如果链表总数是奇数个，返回中间节点，如果链表总数是偶数个，则返回两个中间节点后一个。

关键技巧：快慢指针

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
		//如果是奇数个，返回中间节点；如果是偶数个，返回两个中间节点的后一个
    public ListNode middleNode(ListNode head) {
        ListNode quick = head;
        ListNode slow = head;
        while (quick.next != null && quick.next.next != null){
            quick = quick.next.next;
            slow = slow.next;
        }
        //奇数
        if (quick.next == null){
            return slow;
        } else {
            return slow.next;
        }
    }
}