LeetCode 206. 反转链表 - 迭代和递归（超详细解题思路）

1、题目链接

https://leetcode-cn.com/problems/reverse-linked-list/

2、分析

（1）迭代

使用双指针

直接上代码

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next == null) return head;

        ListNode cur = head;
        ListNode pre = null;
        while(cur != null) {
            ListNode p = cur.next;
            cur.next = pre;
            pre = cur;
            //p  = p.next;
            cur = p;
        }
        
        return pre;
    }
}
           

（2）递归

递归的思路比较难想，见图

上代码：

/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public ListNode reverseList(ListNode head) {
        if(head == null || head.next == null) return head; //边界情况

        ListNode cur = head;
        ListNode next = cur.next;
        ListNode newHead = reverseList(next); //递归，返回反转后的链表的新的头结点
        next.next = cur;
        cur.next = null;
        
        return newHead;
    }
}