题解 | #链表的奇偶重排#

链表的奇偶重排

https://www.nowcoder.com/practice/02bf49ea45cd486daa031614f9bd6fc3

# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None
#
# 代码中的类名、方法名、参数名已经指定,请勿修改,直接返回方法规定的值即可
#
# 
# @param head ListNode类 
# @return ListNode类
#
class Solution:
    def oddEvenList(self , head: ListNode) -> ListNode:
        # write code here
        if not(head and head.next):
            return head
        this = head
        odd = ListNode(this.val)
        this = this.next
        even = ListNode(this.val)
        this_odd = odd
        this_even = even

        this = this.next
        while this:
            this_odd.next = ListNode(this.val)
            this_odd = this_odd.next
            this = this.next
            if this:
                this_even.next = ListNode(this.val)
                this_even = this_even.next
                this = this.next
        this_odd.next = even
        return odd

全部评论

相关推荐

码农索隆:单休一个月少休息4天,一年就是48天,平时节假日,别人3天假期,单休的两天
点赞 评论 收藏
分享
后来123321:别着急,我学院本大二,投了1100份,两个面试,其中一个还是我去线下招聘会投的简历,有时候这东西也得看运气
无实习如何秋招上岸
点赞 评论 收藏
分享
评论
点赞
收藏
分享

创作者周榜

更多
牛客网
牛客网在线编程
牛客网题解
牛客企业服务