技術標籤：leetcode-pythonleetcode連結串列

題目：

給定一個單鏈表 L：L0→L1→…→Ln-1→Ln ，
將其重新排列後變為： L0→Ln→L1→Ln-1→L2→Ln-2→…
你不能只是單純的改變節點內部的值，而是需要實際的進行節點交換。

示例 1:

給定連結串列 1->2->3->4, 重新排列為 1->4->2->3.

示例 2:

給定連結串列 1->2->3->4->5, 重新排列為 1->5->2->4->3.

思路：

找出連結串列中點位置，將後半部分連結串列反轉，然後合併兩條連結串列。

程式碼：

# Definition for singly-linked list.
# class ListNode(object):
#     def __init__(self, val=0, next=None):
#         self.val = val
#         self.next = next
class Solution(object):
    def reorderList(self, head):
        """
        :type head: ListNode
        :rtype: None Do not return anything, modify head in-place instead.
        """
 

        def reverse(head):
            pre = None
            cur = head
            while cur:
                temp = cur.next
                cur.next = pre
                pre = cur
                cur = temp
            return pre
        def merge(head1, head2):
            pre = ListNode(0)
            pre.
 
next = ListNode
            while head1 and head2:
                pre.next = head1
                head1 = head1.next

                pre.next.next = head2
                head2 = head2.next
                pre = pre.next.next
            if head1:
                pre.next = head1
            if head2:
                pre.next = head2

        aim = head
        n = 0
        while aim:
            n += 1
            aim = aim.next

        halflen = (n+1) // 2

        half = head
        pre = ListNode(0)
        pre.next = head

        for i in range(halflen):
            half = half.next
            pre = pre.next
        pre.next = None
        del pre

        half = reverse(half)
        merge(head, half)