力扣 题目96- 不同的二叉搜索树

题目

题解

参考95题的思路 进行排列组合 

即以o(n)=o(i - 1)*o(n-i)的加和--i循环属于[1,n] ; 

也就是需要从1开始然后向上算到n

            for (int j = 1; j <= n;j++ ) {
                int num = 0;
                for (int i = 1; i <= j; i++) {
                   num += res[i - 1] * res[j - i];
                }
                res.push_back(num);
            }

题目

#include<iostream>
#include<vector>
#include<stack>
using namespace std;
class Solution {
public:
    int numTrees(int n) {
        if (n) {
            vector<int>res = {1};
            for (int j = 1; j <= n;j++ ) {
                int num = 0;
                for (int i = 1; i <= j; i++) {
                   num += res[i - 1] * res[j - i];
                }
                res.push_back(num);
            }
            return res.back();
        }
        else return 0;
    }
};



int main() {
    Solution sol;
    int result = sol.numTrees(19);
    cout << result << endl;
}