143. Reorder List

Given a singly linked list L: L0→L1→…→Ln-1→Ln,
reorder it to: L0→Ln→L1→Ln-1→L2→Ln-2→…

You may not modify the values in the list's nodes, only nodes itself may be changed.

Example 1:

Given 1->2->3->4, reorder it to 1->4->2->3.

Example 2:

Given 1->2->3->4->5, reorder it to 1->5->2->4->3.

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, x):
#         self.val = x
#         self.next = None
"""
步骤：
1.小于等于2个节点直接返回
2.找到中间节点（参考：876. Middle of the Linked List）
3.中间节点后面的链表反转（参考：206. Reverse Linked List）  注意：记得断开中间节点后面的链表，不然链表会有环
4.把反转后的链表插入前面的链表
时间复杂度O(n), 空间复杂度O(1)
"""
class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: void Do not return anything, modify head in-place instead.
        """
        if not head or not head.next or not head.next.next:
            return
        
        fast = head
        slow = head
        pre = ListNode(None)
        while fast and fast.next:
            slow = slow.next
            fast = fast.next.next
        
        mid = ListNode(None)
        tmp = slow
        slow = slow.next
        tmp.next = None
        mid.next = slow
        while slow.next:
            tmp = slow.next
            slow.next = slow.next.next
            tmp.next = mid.next
            mid.next = tmp
        
        cur = head
        mid = mid.next
        while mid:
            tmp = cur.next
            tmp1 = mid.next          
            mid.next = cur.next
            cur.next = mid
            cur = tmp
            mid = tmp1