剑指 Offer 06. 从尾到头打印链表

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：

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输入：head = [1,3,2]
输出：[2,3,1]
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#!/usr/bin/env python
# -*- coding: utf-8 -*-
# author：sarizzm time:2021/1/7 0007

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

class Solution:
    def reversePrint(self, head):
        pre = head
        result = []
        if head is None:
            return result
        while pre:
            result.append(pre.val)
            pre = pre.next
        # result.reverse() 
        return result[::-1] # 速度比 result.reverse() 快且节省内存


n1 = [1,2,3,4,5,6]

def generate_node(n1):
    head = ListNode(-1)
    pre = head
    for i in n1:
        pre.next = ListNode(i)
        pre = pre.next
    return head.next

print(Solution().reversePrint(generate_node(n1)))

执行用时：44 ms, 在所有 Python3 提交中击败了77.69%的用户

内存消耗：16.1 MB, 在所有 Python3 提交中击败了33.87%的用户

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/**
 * Definition for singly-linked list.
 * public class ListNode {
 *     int val;
 *     ListNode next;
 *     ListNode(int x) { val = x; }
 * }
 */
class Solution {
    public int[] reversePrint(ListNode head) {
        Stack<ListNode> stack = new Stack<ListNode>();
        ListNode temp = head;
        while (temp != null) {
            stack.push(temp);
            temp = temp.next;
        }
        int size = stack.size();
        int[] print = new int[size];
        for (int i = 0; i < size; i++) {
            print[i] = stack.pop().val;
        }
        return print;
    }
}

//作者：LeetCode-Solution
//链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/solution/mian-shi-ti-06-cong-wei-dao-tou-da-yin-lian-biao-b/