[Swift]LeetCode823. 带因子的二叉树 | Binary Trees With Factors

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

Example 2:

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

---

给出一个含有不重复整数元素的数组，每一个整数均大于 1。

咱们用这些整数来构建二叉树，每一个整数能够使用任意次数。

其中：每一个非叶结点的值应等于它的两个子结点的值的乘积。

知足条件的二叉树一共有多少个？返回的结果应模除 10 ** 9 + 7。

示例 1:

输入: 
输出: 3
解释: 咱们能够获得这些二叉树: A = [2, 4][2], [4], [4, 2, 2]

示例 2:

输入: 
输出: 
解释: 咱们能够获得这些二叉树: .A = [2, 4, 5, 10]7[2], [4], [5], [10], [4, 2, 2], [10, 2, 5], [10, 5, 2]

提示:

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

---

Runtime: 132 ms

Memory Usage: 19.3 MB

 1 class Solution {
 2     func numFactoredBinaryTrees(_ A: [Int]) -> Int {
 3         var res:Int = 0
 4         var M:Int = Int(1e9 + 7)
 5         var dp:[Int:Int] = [Int:Int]()
 6         var A = A.sorted(by:<)
 7         for i in 0..<A.count
 8         {
 9             dp[A[i]] = 1
10             for j in 0..<i
11             {
12                 if A[i] % A[j] == 0 && dp[A[i] / A[j]] != nil
13                 {
14                     dp[A[i]] = dp[A[i],default:0] + dp[A[j],default:0] * dp[Int(A[i] / A[j]),default:0] % M
15                 }
16             }
17         }
18         for (key,val) in dp 
19         {
20             res = (res + val) % M
21         }
22         return res
23     }
24 }