Article Directory
1. Title
Given an array containing unique integer elements, each integer is greater than 1.
We use these integers to build a binary tree, and each integer can be used any number of times .
Among them: the value of each non-leaf node should be equal to the product of the values of its two child nodes .
How many binary trees meet the conditions? The returned result should be modulo 10 ** 9 + 7.
示例 1:
输入: A = [2, 4]
输出: 3
解释: 我们可以得到这些二叉树: [2], [4], [4, 2, 2]
示例 2:
输入: A = [2, 4, 5, 10]
输出: 7
解释: 我们可以得到这些二叉树: [2], [4], [5], [10],
[4, 2, 2], [10, 2, 5], [10, 5, 2].
提示:
1 <= A.length <= 1000.
2 <= A[i] <= 10 ^ 9.
Source: LeetCode (LeetCode)
Link: https://leetcode-cn.com/problems/binary-trees-with-factors
Copyright is owned by LeetCode . For commercial reprints, please contact the official authorization. For non-commercial reprints, please indicate the source.
2. Problem solving
- d p [ i ] dp[i] d p [ i ] meansA [i] A[i]A [ i ] is the number of root binary trees
- A [ i ] > 1 A[i]>1 A[i]>1 The higher the value, the greater the value of the root, sort the array, and hash themap mapm a p record position
- p = A [ i ] ∗ A [ j ] p= A[i]*A[j] p=A[i]∗A [ j ] in the array,i ≠ j, dp [map [p]] + = dp [i] ∗ dp [j] ∗ 2 i \neq j, dp[map[p]] += dp[i] *dp[j]*2i=j,dp[map[p]]+=dp[i]∗dp[j]∗2. Left and right nodes can be interchanged
*2 - p = A [ i ] ∗ A [ j ] p= A[i]*A[j] p=A[i]∗A[j] 在数组中， i = j , d p [ m a p [ p ] ] + = d p [ i ] ∗ d p [ j ] i = j, dp[map[p]] += dp[i]*dp[j] i=j,dp[map[p]]+=dp[i]∗dp[j]
- The final answer is sum (dp [i])% mod sum(dp[i])\%modsum(dp[i])%mod
class Solution {
public:
int numFactoredBinaryTrees(vector<int>& A) {
int n = A.size(), i, j;
if(n == 1) return 1;
vector<long long> dp(n, 1);
sort(A.begin(), A.end());
unordered_map<int,int> m;
for(i = 0; i < n; i++)
m[A[i]] = i;
long long ans = 0, p, mod = 1e9+7;
for(i = 0; i < n; ++i)
{
for(j = 0; j <= i; ++j)
{
p = (long long)A[j]*A[i];
if(p > int(1e9))
break;
if(m.find(p) != m.end())
{
if(i != j)
dp[m[p]] += (2*dp[i]*dp[j])%mod;
else
dp[m[p]] += (dp[i]*dp[j])%mod;
}
}
}
for(int i = 0; i < n; i++)
ans = (ans+dp[i])%mod;
return ans;
}
};
52 ms 9.5 MB
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