LeetCode206 | Reverse Linked List (Easy)

题目地址(206. 反转链表)

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

题目描述

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

前置知识

• 构造函数

// Definition for singly-linked list.
public class ListNode {
    int val;
    ListNode next;
    ListNode() {}
    ListNode(int val) { this.val = val; }
    ListNode(int val, ListNode next) { this.val = val; this.next = next; }
}

思路

• 思路一：头插法（迭代）
• 思路二：递归

关键点

• 可以尝试画图理清节点与节点之间的关系

a = b;
a = b.next;
a.next = b;
a.next = b.next;

代码

• 语言支持：Java

Java Code:

方法一：头插法（迭代）

class Solution {
    public ListNode reverseList(ListNode head) {
        ListNode newHead = new ListNode(-1);
        while (head != null) {
            ListNode next = head.next;
            head.next = newHead.next;
            newHead.next = head;
            head = next;
        }
        return newHead.next;
    }
}

复杂度分析

令 n 为数组长度。

• 时间复杂度：$O(n)$
• 空间复杂度：$O(n)$

起始状态

1.定义newHead

ListNode newHead = new ListNode(-1);

2.定义next

next = head.next;

3.第一个节点指向null

head.next = newHead.next;

4.newHead指向头节点

newHead.next = head;

5.head往后移动

head = next;
next = head.next;

全过程

public ListNode reverseList(ListNode head) {
    ListNode newHead = new ListNode(-1);
    while (head != null) {
        ListNode next = head.next;
        head.next = newHead.next;
        newHead.next = head;
        head = next;
    }
    return newHead.next;
}

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方法二：递归

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

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