day3 力扣

链表反转

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        ListNode newHead = reverseList(head.next);
        head.next.next = head;
        head.next = null;
        return newHead;
    }
}

讲解链接【入门算法题】链表反转-迭代_哔哩哔哩_bilibili