首页 > 其他 > 详细

[LeetCode] Binary Trees With Factors 带因子的二叉树

时间:2019-01-05 00:37:44      阅读:223      评论:0      收藏:0      [点我收藏+]

 

Given an array of unique integers, each integer is strictly greater than 1.

We make a binary tree using these integers and each number may be used for any number of times.

Each non-leaf node‘s value should be equal to the product of the values of it‘s children.

How many binary trees can we make?  Return the answer modulo 10 ** 9 + 7.

Example 1:

Input: A = [2, 4]
Output: 3
Explanation: We can make these trees: [2], [4], [4, 2, 2]

Example 2:

Input: A = [2, 4, 5, 10]
Output: 7
Explanation: We can make these trees: [2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2].

 

Note:

  1. 1 <= A.length <= 1000.
  2. 2 <= A[i] <= 10 ^ 9.
 

s

 

[LeetCode] Binary Trees With Factors 带因子的二叉树

原文:https://www.cnblogs.com/grandyang/p/10223109.html

(0)
(0)
   
举报
评论 一句话评论(0
关于我们 - 联系我们 - 留言反馈 - 联系我们:wmxa8@hotmail.com
© 2014 bubuko.com 版权所有
打开技术之扣,分享程序人生!