代码随想录算法训练营第40天

class Solution {
    public int integerBreak(int n) {
        int[] dp = new int[n + 1];
        dp[2] = 1;
        for (int i = 3; i <= n; i++) {
            for (int j = 1; j < i; j++) {
                dp[i] = Math.max(dp[i], Math.max(j*(i-j), j*dp[i-j]));
            }
        }
        // System.out.println(Arrays.toString(dp));
        return dp[n];
    }
}

class Solution {
    public int numTrees(int n) {
        // dp[i]: 长度为1~i的数可以构成几种二叉搜索树？
        int[] dp = new int[n + 1];
        dp[0] = 1;
        dp[1] = 1;
        // dp[2] = 2;
        // dp[3] = 5;
        for (int i = 2; i <= n; i++) {
            // 以j为当前root节点，左右子树的节点数分别为j-1和i-j
            for (int j = 1; j <= i; j++) {
                dp[i] += dp[j - 1] * dp[i - j];
                // dp[i] = Math.max(dp[i], Math.max());
            }
        }
        // System.out.println(Arrays.toString(dp));
        return dp[n];
    }
}