343. 整数拆分

数组表示一个集合，由于可能出现两个或多个相乘，所以递推的公式是 dp[i] = max(dp[i], j * dp[i-j]); dp[i] = max(dp[i], j * (i-j));

// dp[i] = dp[x] * dp[i-x]
class Solution {
public:
    int integerBreak(int n) {
        vector<int> dp(n + 1);
        if (n == 2) return 1;
        if (n == 3) return 2;
        dp[1] = 1;
        dp[2] = 1;
        dp[3] = 2;
        for (int i = 4; i <= n; i ++)
        {
            for (int j = 2; j <= i - 2; j ++)
            {
                dp[i] = max(dp[i], j * dp[i-j]);
                dp[i] = max(dp[i], j * (i-j));
            }
        }
        return dp[n];
    }
};

96. 不同的二叉搜索树

这题比较难，我觉得再碰到也不会写。

class Solution {
public:
    int numTrees(int n) {
        vector<int>dp(n + 1);
        dp[0] = 1;
        for (int i = 1; i <= n; i ++)
            for (int j = 1; j <= i; j ++)
                dp[i] += dp[j-1] * dp[i-j];

        return dp[n];
    }
};