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)O(n)O(n)
空间复杂度：O(n)O(n)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;
}

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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LeetCode206 | Reverse Linked List (Easy)

题目地址(206. 反转链表)

题目描述

前置知识

思路

关键点

代码

方法一：头插法（迭代）

方法二：递归

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