LeetCode *328. Odd Even Linked List (Python Solution)

题目描述

Given a singly linked list, group all odd nodes together followed by the even nodes. Please note here we are talking about the node number and not the value in the nodes.

You should try to do it in place. The program should run in O(1) space complexity and O(nodes) time complexity.

给定单链表,将所有奇数节点组合在一起,然后是偶数节点。 请注意,这里我们讨论的是节点编号,而不是节点中的值。

你应该尝试到位。 该程序应该以O(1)空间复杂度和O(节点)时间复杂度运行。

Example 1:

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

Example 2:

Input: 2->1->3->5->6->4->7->NULL
Output: 2->3->6->7->1->5->4->NULL

Python Solution

分析: 将链表分成奇偶两条然后进行拼接即可,重点是边界判断

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

class Solution:
    def oddEvenList(self, head: ListNode) -> ListNode:
        if not head:
            return None
        odd, even = head, head.next
        eventmp = even
        while even and even.next:
            odd.next = even.next
            odd = odd.next
            even.next = odd.next
            even = even.next
        odd.next = eventmp
        return head
评论
添加红包

请填写红包祝福语或标题

红包个数最小为10个

红包金额最低5元

当前余额3.43前往充值 >
需支付:10.00
成就一亿技术人!
领取后你会自动成为博主和红包主的粉丝 规则
hope_wisdom
发出的红包
实付
使用余额支付
点击重新获取
扫码支付
钱包余额 0

抵扣说明:

1.余额是钱包充值的虚拟货币,按照1:1的比例进行支付金额的抵扣。
2.余额无法直接购买下载,可以购买VIP、付费专栏及课程。

余额充值