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:
            return None
        first, second = head, head.next
        tmp = second
        while second and second.next:
            first.next = first.next.next
            second.next = second.next.next
            first = first.next
            second = second.next
        first.next = tmp
        return head

当first second指针往后移动时，原先的基数链接会被替换为偶数链接，因为next.next