Unique Binary Search Trees

class Solution {
public:
    int numTrees(int n) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        vector<int> num;
        num.push_back(1);
        num.push_back(1);
        int temp = 1;
        for(int i = 2;i<=n;i++)
        {
            temp = 0;
            for(int j = 1;j<=i;j++)
            {
                temp += num[j-1] * num[i-j];
            }
            num.push_back(temp);
        }
        return temp;
    }
};

唯一搜索树的个数。leetcode上的题，大概思想是递归，但是用数组存储已计算的值可以减少很多重复计算，就像斐波那契数列那样，如果用递归则发费太多的时间。