【链表】876.链表的中间结点

题目

法1：快慢指针

class Solution {
    public ListNode middleNode(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode dummy = new ListNode(-1);
        dummy.next = head;
        ListNode fast = dummy, slow = dummy;
        while (fast.next != null && fast.next.next != null) {
            fast = fast.next.next;
            slow = slow.next;
        }

        return slow.next;
    }
}