leetcode 823 带因子的二叉树

用动态规划

如果两个节点值不同,要乘2，因为两个节点可以互换位置

dp[i] = dp[left] * dp[right] * 2

如果相同

dp[i] = dp[left] * dp[right]

class Solution {
    public int numFactoredBinaryTrees(int[] arr) {
        Arrays.sort(arr);
        int n = arr.length;
        long[] dp = new long[n];
        long res = 0, mod = 1000000007;
        for (int i = 0; i < n; i++) {
            dp[i] = 1;
            for (int left = 0, right = i - 1; left <= right; left++) {
                while (right >= left && (long) arr[left] * arr[right] > arr[i]) {
                    right--;
                }
                if (right >= left && (long) arr[left] * arr[right] == arr[i]) {
                    if (right != left) {
                        dp[i] = (dp[i] + dp[left] * dp[right] * 2) % mod;
                    } else {
                        dp[i] = (dp[i] + dp[left] * dp[right]) % mod;
                    }
                }
            }
            res = (res + dp[i]) % mod;
        }
        return (int) res;
    }
}