Java 详解分析链表的中间节点

class Solution {
    public ListNode middleNode(ListNode head) {
        if (head == null) {
            return null;
        }
        ListNode fast = head;
        ListNode slow = head;
        while (fast != null && fast.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }
        return slow;
 
    }
}