【剑指Offer系列06】从尾到头打印链表

文章目录

题目

输入一个链表的头节点,从尾到头反过来返回每个节点的值(用数组返回)。

示例 1:
输入:head = [1,3,2]
输出:[2,3,1]

限制:
0 <= 链表长度 <= 10000

代码

Python
# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

# 思路:
# 打印且不改变链表结构,采用辅助栈或递归
# 复杂度:
# O(N)
class Solution:
    def reversePrint(self, head: ListNode) -> List[int]:
        stack = []
        while head:
            stack.append(head.val)
            head = head.next
        return stack[::-1]
C++
/**
 * 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> res;
        while (head) {
            res.push_back(head->val);
            head = head->next;
        }
        reverse(res.begin(), res.end());
        return res;
    }
};