剑指 Offer 24. 反转链表
mengjian-github opened this issue · 2 comments
mengjian-github commented
定义一个函数,输入一个链表的头节点,反转该链表并输出反转后链表的头节点。
示例:
输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL
来源:力扣(LeetCode)
链接:https://leetcode.cn/problems/fan-zhuan-lian-biao-lcof
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mengjian-github commented
思路1:创建一个新的链表返回:
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var reverseList = function(head) {
let result = null;
while(head) {
const n = new ListNode(head.val);
n.next = result;
result = n;
head = head.next;
}
return result;
};mengjian-github commented
思路2:可以不创建一个链表,直接遍历每个节点,改变next就好,不过要保存一下上一个节点:
/**
* Definition for singly-linked list.
* function ListNode(val) {
* this.val = val;
* this.next = null;
* }
*/
/**
* @param {ListNode} head
* @return {ListNode}
*/
var reverseList = function(head) {
let prev = null;
let cur = null;
while(head) {
cur = head;
head = head.next;
cur.next = prev;
prev = cur;
}
return prev;
};