reverse-linked-list.md

206. Reverse Linked List - 反转链表

Tags - 题目标签

Description - 题目描述

EN:

Given the head of a singly linked list, reverse the list, and return the reversed list.

 

Example 1:

Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2:

Input: head = [1,2]
Output: [2,1]

Example 3:

Input: head = []
Output: []

 

Constraints:

• The number of nodes in the list is the range [0, 5000].
• -5000 <= Node.val <= 5000

 

Follow up: A linked list can be reversed either iteratively or recursively. Could you implement both?

ZH-CN:

给你单链表的头节点 head ，请你反转链表，并返回反转后的链表。

 

示例 1：

输入：head = [1,2,3,4,5]
输出：[5,4,3,2,1]

示例 2：

输入：head = [1,2]
输出：[2,1]

示例 3：

输入：head = []
输出：[]

 

提示：

• 链表中节点的数目范围是 [0, 5000]
• -5000 <= Node.val <= 5000

 

进阶：链表可以选用迭代或递归方式完成反转。你能否用两种方法解决这道题？

Link - 题目链接

LeetCode - LeetCode-CN

Latest Accepted Submissions - 最近一次 AC 的提交

Language	Runtime	Memory	Submission Time
javascript	72 ms	34.2 MB	2020/03/23 12:45

/**
 * Definition for singly-linked list.
 * function ListNode(val) {
 *     this.val = val;
 *     this.next = null;
 * }
 */
/**
 * @param {ListNode} head
 * @return {ListNode}
 */
var reverseList = function(head) {
    var preNode, curNode, nextNode;
    if (!head) {
        return null;
    }

    if(head.next === null) {
        return head;
    }

    preNode = head, curNode = head.next;
    head.next = null;
    while(curNode.next) {
        nextNode = curNode.next;
        curNode.next = preNode;
        preNode = curNode;
        curNode = nextNode;
    }
    curNode.next = preNode;
    return curNode;
};

My Notes - 我的笔记

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