剑指 Offer 06. 从尾到头打印链表

剑指 Offer 06. 从尾到头打印链表

链接：https://leetcode-cn.com/problems/cong-wei-dao-tou-da-yin-lian-biao-lcof/

 

数组简单应用，依次存储链表各个节点的值，然后反转就好了

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

class Solution:
    def reversePrint(self, head: ListNode) -> List[int]:
        stack = []
        while head:
            stack.append(head.val)
            head = head.next
        return stack[::-1]