[Algorithmic programming C++ python] singly linked list reverse order output

Topic description

Enter a linked list and print the value of each node in the linked list from end to beginning.

 

The following methods are only functional and may not be the best. Tested on Niuke.com,

C++：3ms 480k

Python：23ms 5732k

/**
*  struct ListNode {
*        int val;
*        struct ListNode *next;
*        ListNode(int x) :
*              val(x), next(NULL) {
*        }
*  };
*/
class Solution {
public:
    vector<int> printListFromTailToHead(ListNode* head) {
        vector<int> vec_output;
        if (head == NULL){return vec_output;}
        do{
            vec_output.push_back(head->val);
            head = head->next;
        }while(head!=NULL);
        reverse(vec_output.begin(),vec_output.end());
        return vec_output;
    }
};

Python：

# -*- coding:utf-8 -*-
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
     #Return the sequence of list values ​​from tail to head, eg [1,2,3] 
    def printListFromTailToHead(self, listNode):
        list_val = []
        while(True):
            if listNode == None: return [] 
            list_val.append(listNode.val)
            if (listNode.next):
                listNode = listNode.next
            else:
                break
        return list_val[::-1]