NYUer | LeetCode206 Reverse Linked List-优快云博客

本文链接：https://blog.youkuaiyun.com/Moses_SU/article/details/127569475

该博客围绕LeetCode206反转单链表问题展开。介绍了问题描述、示例及约束条件，分析中提出用两个索引操作节点实现反转，给出双索引解法，还提到链表反转可迭代或递归实现，并给出递归版本，希望能为解题提供启发。

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LeetCode206 Reverse Linked List

Author: Stefan Su
Create time: 2022-10-28 00:49:07
Location: New York City, NY, USA

Description `Easy`

Given the head of a singly linked list, reverse the list, and return the reversed list.

Example 1

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Input: head = [1,2,3,4,5]
Output: [5,4,3,2,1]

Example 2

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Input: head = [1,2]
Output: [2,1]

Example 3

Input: head = []
Output: []

Constrains

The number of nodes in the list is the range [0, 5000].
-5000 <= Node.val <= 5000

Analysis

Use two indices, cur and prev to indicate the two nodes to be operated. In each while loop, use temp variable to store cur.next first. And then, set cur.next to prev to complete reverse. Finally, update prev to cur and cur to temp. Attention, the order is really important here. We stop while loop until cur pointer points to a null pointer.

Solution

Two indices version.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """

        # Two index version
        cur = head
        prev = None

        while cur != None:
            temp = cur.next
            cur.next = prev
            prev = cur
            cur = temp
        
        return prev

Advance

A linked list can be reversed either iteratively or recursively.

Recursion version

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reverseList(self, head):
        """
        :type head: ListNode
        :rtype: ListNode
        """
        # recursion version
        def reverse(cur, prev):
            if cur == None:
                return prev

            temp = cur.next
            cur.next = prev
            return reverse(temp, cur)

        return reverse(head, None)

    # can also write the reserve function outside of reserveList, but add self to call it
    # def reverse(self, cur, prev):
    #     if cur == None:
    #         return prev

    #     temp = cur.next
    #     cur.next = prev
    #     return self.reverse(temp, cur)