【LeetCode 823】 Binary Trees With Factors_binary trees with factors leetcode-优快云博客

本文链接：https://blog.youkuaiyun.com/iCode_girl/article/details/104524912

本文探讨了一道关于二叉树构造的问题，利用动态规划方法解决数组中元素构成二叉树的不同方式数量，通过实例解析算法思路并提供C++实现代码。

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题目描述

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

思路

动态规划，注意每个单独的数字也算构成一棵合法的二叉树。那么，对于数字 n 来说，mp[n] += mp[a] * mp[b]，每个数字初始化为1。也就是对当前的数字的合法数，子问题是数组中所有能组成当前数字的合法数乘积之和。遍历数组，对于每个数字，遍历比它小的数字，如果能和其它数字乘积，加之。不同的数字乘积正好能找到两遍，相同的只能找到一遍。

代码

class Solution {
public:
    int numFactoredBinaryTrees(vector<int>& A) {
        int n = A.size();
        int MOD = 1e9+7;
        
        sort(A.begin(), A.end());
        map<int, long long> mp;
        for (int i=0; i<n; ++i) {
            mp[A[i]] = 1;
        }
        
        int res = 0;
        for (int i=0; i<n; ++i) {
            int num1 = A[i], num2;
            for (int j=0; j<i; ++j) {
                num2 = A[j];
                if (num1%num2 == 0 && mp[num1/num2]) {
                    mp[num1] += mp[num2]*mp[num1/num2];
                    mp[num1] %= MOD;
                }
            }
            res = (res + mp[num1]) % MOD;
        }
        
        return res;
    }
};