【Leetcode】143—Reorder List

一、题目描述

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：

给定链表 1->2->3->4, 重新排列为 1->4->2->3.

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

二、代码实现

三步：1.找到中点 2.逆转后半部分 3.拼接前半部分和逆转过的后半部分

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

class Solution(object):
    def reverseList(self, cur):
        pre = None
        while cur:
            last = cur.next
            cur.next = pre
            pre = cur
            cur = last
        return pre
    
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """
        if head == None or head.next == None: return head
        
        fast = head
        slow = head
        
        while fast.next and fast.next.next:
            fast = fast.next.next
            slow = slow.next
            
        newHead = self.reverseList(slow.next)
        slow.next = None
        
        p = head
        while newHead:
            orinext = p.next
            p.next = newHead
            newHead = newHead.next
            p = p.next
            p.next = orinext
            p = p.next