一、题目内容

二、题解

解法1：

思路

要求空间复杂度O(1)，空间复杂度O(n)，所以只能递归或者尾插遍历
pre,curr,next

代码

//尾插遍历
public 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;
    }
}
//递归
public 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;
    }
}