LeetCode | 0206. Reverse Linked List反转链表【Python】

LeetCode 0206. Reverse Linked List反转链表【Easy】【Python】【链表】

Problem

LeetCode

Reverse a singly linked list.

Example:

Input: 1->2->3->4->5->NULL
Output: 5->4->3->2->1->NULL

Follow up:

A linked list can be reversed either iteratively or recursively. Could you implement both?

问题

力扣

反转一个单链表。

示例:

输入: 1->2->3->4->5->NULL
输出: 5->4->3->2->1->NULL

进阶:
你可以迭代或递归地反转链表。你能否用两种方法解决这道题？

思路

解法一

栈

普通数组模拟栈实现反转链表

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

Python3代码

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

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        # solution one: List
        pos = head
        newList = []
        while pos:
            newList.insert(0, pos.val)
            pos = pos.next

        pos = head
        for _ in newList:
            pos.val = _
            pos = pos.next
        return head

解法二

双指针

用 cur 和 pre 双指针实现反转链表。
可以动手在纸上画画，就能理解过程了。

时间复杂度: O(n)，n 为链表长度。
空间复杂度: O(1)

Python3代码

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

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        # solution two: two point
        cur, pre = None, head
        while pre:
            pos = pre.next
            pre.next = cur
            cur = pre
            pre = pos
        return cur

解法三

递归

直接利用递归实现反转链表。

时间复杂度: O(n)，n 为链表长度。
空间复杂度: O(1)

Python3代码

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

class Solution:
    def reverseList(self, head: ListNode) -> ListNode:
        # solution three: recursion
        if not head or not head.next:
            return head

        pos = head.next
        newList = self.reverseList(pos)

        head.next = None
        pos.next = head
        return newList

代码地址

GitHub链接