链表的中间节点

leetcode对应题目

代码

快慢指针

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode() {}
 *     ListNode(int val) { this.val = val; }
 *     ListNode(int val, ListNode next) { this.val = val; this.next = next; }
 * }
 */
class Solution {
    public ListNode middleNode(ListNode head) {

        //快指针
        ListNode fast = head;
        //慢指针
        ListNode slow = head;

        //快指针每次走两步，慢指针每次走一步。当链表节点数为奇数时，fast.next==null；当链表节点数为偶数时，fast==null。
        while(fast!=null&&fast.next!=null){
            fast = fast.next.next;
            slow = slow.next;
        }

        //返回从中间节点开始的序列化链表
        return slow;

    }
}
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图解