题目描述

思路 & 代码

• 一次遍历：翻转链表 + 记录长度
• 二次遍历：遍历翻转后的链表，更新数组
• O(n) & O(1)

/*** Definition for singly-linked list.* public class ListNode {*     int val;*     ListNode next;*     ListNode(int x) { val = x; }* }*/
class Solution {public int[] reversePrint(ListNode head) {// O(n) & O(1)// 一次遍历：翻转链表 + 记录长度ListNode now = head;ListNode next = null;ListNode temp = null;int len = 0;while(now != null){temp = now.next;now.next = next;next = now;now = temp;len++;}// 二次遍历：赋值int[] ans = new int[len];for(int i = 0; i < len; i++, next = next.next){ans[i] = next.val;}return ans;}
}

二刷

• 其实不用翻转也行= =，从后往前填即可

class Solution {public int[] reversePrint(ListNode head) {int len = 0;ListNode temp = head;while(temp != null) {len++;temp = temp.next;}int[] ans = new int[len];temp = head;for(int i = len - 1; i >= 0; i--, temp = temp.next) {ans[i] = temp.val;}return ans;}
}