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题目

思路分析

这个题目的思想是比较巧妙的，拆解来看都是我们熟悉的三种基本的链表操作

链表记得画图！

实现代码

# Definition for singly-linked list.# class ListNode(object):#     def __init__(self, x):#         self.val = x#         self.next = Noneclass Solution(object):    def reorderList(self, head):        """        :type head: ListNode        :rtype: void Do not return anything, modify head in-place instead.        """        # 思路一开始就错了，自己是个太暴力的人，但是机器不应该这样        # 0个1个两个节点不需要进行这个操作        if head and head.next and head.next.next:            # 将链表一分为二            fast = slow = head            while fast and fast.next:                fast = fast.next.next                slow = slow.next            head1,head2 = head,slow.next            slow.next = None            # 翻转后半部分链表            dummy = ListNode(0)            dummy.next = head2            p = head2.next            head2.next = None            while p:                tmp,p = p,p.next                tmp.next = dummy.next                dummy.next = tmp            head2 = dummy.next            # 重新将两部分连接起来            p1 = head1;             p2 = head2            while p2:                tmp1 = p1.next; tmp2 = p2.next                p1.next = p2; p2.next = tmp1                p1 = tmp1; p2 = tmp2