143. 重排链表（中等，链）（12.25）

给定一个单链表 L：L0→L1→…→Ln-1→Ln ，
将其重新排列后变为： L0→Ln→L1→Ln-1→L2→Ln-2→…

你不能只是单纯的改变节点内部的值，而是需要实际的进行节点交换。

示例 1:

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

示例 2:

给定链表 1->2->3->4->5, 重新排列为 1->5->2->4->3.

# Definition for singly-linked list.
# class ListNode:
#     def __init__(self, x):
#         self.val = x
#         self.next = None

class Solution:
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: void Do not return anything, modify head in-place instead.
        """
        if not head:
            return None
        slow=head
        fast=head
        while fast and fast.next:
            slow=slow.next
            fast=fast.next.next
        head1=head
        head2=slow.next
        slow.next = None   #不是很清楚这句话的意思
        #反转第二条链
        cur, pre = head2, None
        while cur:
            nex = cur.next
            cur.next = pre
            pre = cur
            cur = nex
        #交叉
        cur1, cur2 = head1, pre
        while cur2:
            nex1, nex2 = cur1.next, cur2.next
            cur1.next = cur2
            cur2.next = nex1
            cur1, cur2 = nex1, nex2

执行用时: 116 ms, 在Reorder List的Python3提交中击败了95.60% 的用户