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
        if not head or not head.next:
            return head
        oddList = ListNode(0)
        evenList = ListNode(0)
        count = 1
        cur = head
        o = oddList
        e = evenList
        while cur:
            if count % 2 != 0:
                o.next = ListNode(cur.val)
                o = o.next
            else:
                e.next = ListNode(cur.val)
                e = e.next
            cur = cur.next
            count += 1
        o.next = evenList.next
        return oddList.next