边工作边刷题：70天一遍leetcode: day 10-2

Reorder List

要点：算法直观不多讲，这题就是实现有些细节要注意

• 同样是前面提到的slow和fast的算法最终slow落在奇数结点中点或者偶数结点的中间右侧。所以第二步reverse的起点是slow.next：奇数很明显，偶数个，右面list的第一个结点是reverse后的最后一个结点，是不移动的
• reverse前一定不要忘了把前后两个list分开(即set slow.next=None)，否则后一个list在reverse后是没有null end的

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

class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: void Do not return anything, modify head in-place instead.
        """
        if not head: return
        fast = slow = head
        while fast:
            fast=fast.next
            if fast:
                fast=fast.next
                slow=slow.next
        #print fast.val, slow.val
        
        if not slow.next: return
        cur = slow.next
        pre = None
        slow.next = None
        while cur:
            next = cur.next
            cur.next = pre
            pre = cur
            cur = next
        first = head
        second = pre
        print first.val,second.val
        while first and second:
            next = second.next
            #print next.val
            second.next = first.next
            first.next = second
            first = second.next
            second = next