题解 | #链表的奇偶重排#

解题思路：

1. 设置两个列表，分别保存链表中的奇数结点和偶数结点
2. 直接根据链表中的值创建新节点，直到完成实现链表的奇偶重排

#     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:
            return head
        res = ListNode(-1)
        lst1 = []
        lst2 = []
        p = head
        count = 1
        while p:
            if count % 2 != 0:
                lst1.append(p.val)
            else:
                lst2.append(p.val)
            count += 1
            p = p.next
        print(lst1, lst2)
        q = res
        for i in range(len(lst1)):
            q.next = ListNode(lst1[i])
            q = q.next
        for i in range(len(lst2)):
            q.next = ListNode(lst2[i])
            q = q.next
        return res.next