LeetCode | 面试题06. 从尾到头打印链表剑指OfferPython

Posted 2021-03-05 wonz

LeetCode 面试题06. 从尾到头打印链表【剑指Offer】【Easy】【Python】【链表】

问题

力扣

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

思路

解法一

reverse函数

时间复杂度: O(n)，n为 head 链表长度。
空间复杂度: O(n)，n为 head 链表长度。

Python3代码

# 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]:
        # solution one: reverse
        res = []
        while head:
            res.append(head.val)
            head = head.next
        res.reverse()
        return res

解法二

栈

时间复杂度: O(n)，n为 head 链表长度。
空间复杂度: O(n)，n为 head 链表长度。

Python3代码

# 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]:
        # solution two: 栈
        stack = []
        while head:  # push
            stack.append(head.val)
            head = head.next
        res = []
        while stack:  # pop
            res.append(stack.pop())
        return res

解法三

递归

Python3代码

# 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]:
        # solution three: 递归
        return self.reversePrint(head.next) + [head.val] if head else []

代码地址

GitHub链接