一、题目内容

二、题解

解法1：

思路

遍历尾插法。建立 pre 节点、临时存储 curr.next 节点。时间On，空间O1

代码

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode pre = null;
        ListNode curr = head;
        while (curr != null) {
            ListNode next = curr.next;
            curr.next = pre;
            pre = curr;
            curr = next;
        }
        return pre;
    }
}

解法2：

思路

递归

代码

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;
    }
}

解法3：

思路

堆栈

代码

class Solution {
    public ListNode reverseList(ListNode head) {
        if (head == null || head.next == null) {
            return head;
        }
        Stack<ListNode> stack = new Stack<ListNode>();
        //存进去
        while (head != null) {
            stack.push(head);
            head = head.next;
        }
        //取出来
        ListNode curr = stack.pop();
        //记录新head
        ListNode newHead = curr;
        //清空原来存储的不对的next
        curr.next = null;
        while (!stack.isEmpty()) {
            curr.next = stack.pop();
            curr = curr.next;
            curr.next = null;
        }
        return newHead;
    }
}