206. 反转链表
问题描述

问题思路
- **利用外部空间:**将所给链表存到ArryList里面或者是新的链表里面,然后再反转动态数组就可以了。
- 快慢指针
- 递归解法
代码实现
js
var reverseList = function(head) {
let prev = null;
let curr = head;
while (curr) {
const next = curr.next;
curr.next = prev;
prev = curr;
curr = next;
}
return prev;
};
递归实现
if (head == null || head.next == null) return head;
ListNode newHead = reverseList(head.next);
head.next.next = head;
head.next = null;
return newHead;
快慢指针
class Solution {
public ListNode reverseList(ListNode head) {
if (head == null || head.next == null) return head;
ListNode newHead = null;
while (head != null){
ListNode tmp = head.next;
head.next = newHead;
newHead = head;
head = tmp;
}
return newHead;
}
}
树下留言
LET’S TALK文字是一次相遇。很高兴听到你的声音。