class Solution {
public:
    int numSquares(int n) {
        vector<int> f(n + 1);
        for (int i = 1; i <= n; i++) {
            int minn = INT_MAX;
            for (int j = 1; j * j <= i; j++) {
                minn = min(minn, f[i - j * j]);   //上一次的 & 当前数可以找到一个新的更大的平方
            }
            f[i] = minn + 1;     
        }
        return f[n];
    }
};