Sword finger offer 06 Print linked list from end to end

subject ：

Enter the head node of a linked list , Return the value of each node from the end to the end （ Return with array ）.

Example 1：

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

Limit ：

0 <= Chain length <= 10000

source ： Power button （LeetCode）
link ：https://leetcode.cn/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof
Copyright belongs to the network . For commercial reprint, please contact the official authority , Non-commercial reprint please indicate the source .

Ideas ：

I've been relatively idle recently , Think again, sword finger . This is a simple question , The common idea is to go through it list Deposit in vector, Again reverse vector; Or on the spot reverse list, Re deposit vector; Or deposit stack Again pop come out , These differences are not big . Wrote the simplest version of the code 1. And then when I look at other people's answers , I saw a version of powerful recursion , Put it in the code 2 了 . Indeed, this problem is very clear in the case of a small amount of data , But I don't usually think of . here base case Namely head It's empty , Go straight back to an empty vector, And every time we get vector Are returned by the next level of recursion , Because it's in reverse order , We will return to the current layer after the next layer returns val In the vector The end of , This returns vector It's in reverse order .

Code 1：

/**

 * Definition for singly-linked list.

 * struct ListNode {

 *     int val;

 *     ListNode *next;

 *     ListNode(int x) : val(x), next(NULL) {}

 * };

 */

class Solution {

public:

    vector<int> reversePrint(ListNode* head) {

        vector<int> record;

        while (head) {

            record.push_back(head->val);

            head = head->next;

        }

        reverse(record.begin(), record.end());

        return record;

    }

};

Code 2：