96. Unique Binary Search Trees

96. Unique Binary Search Trees

动态规划

class Solution {
public:
    int numTrees(int n) {
        int *dp = new int[n + 1]{ 0 };
        dp[0] = dp[1] = 1;
        for(int i = 2; i <= n; ++i)
        {
            for(int j = 1; j <= i; ++j)
                dp[i] += dp[j - 1] * dp[i - j];
        }
        return dp[n];
    }
};

抄了discussion

将问题分解为每个小小的n会生成多少的树，最后左树数 * 右树数

review

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.back();
    }
};