leetcode 823. Binary Trees With Factors

leetcode 823. Binary Trees With Factors

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原题地址：https://leetcode.com/problems/binary-trees-with-factors/

题目

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.

python代码

class Solution:
    def numFactoredBinaryTrees(self, A):
        """
        :type A: List[int]
        :rtype: int
        """
        A = list(set(A))
        A.sort() 
        s = {}
        for i in range(len(A)):
            s[A[i]] = 1
            for j in range(i):
                if A[i] % A[j] == 0 and A[i] // A[j] in s.keys():
                    s[A[i]] += s[A[j]] * s[A[i] // A[j]]

        return sum(s.values()) % (10**9 + 7)

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