题解 | #链表的奇偶重排#

# 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
        num = 1
        odd = ListNode(0)
        odd_dummy = odd
        even = ListNode(0)
        even_dummy = even
        while head:
            if(num % 2):
                odd.next = ListNode(head.val)
                odd = odd.next
            else:
                even.next = ListNode(head.val)
                even = even.next
            num += 1
            head = head.next
        odd.next = even_dummy.next
        return odd_dummy.next