leetcode279.完全平方数

1.动态规划
状态转移方程：dp[i] = min(dp[i], dp[ i - j * j]+1)，其中j是满足：j*j < i，的平方数的平方根。
其中dp[i]表示和为i的数最少由几个平方数组成

int numSquares(int n) 
    {
        if (n <= 0)
        {
            return 0;
        }
        
        // cntPerfectSquares[i] = the least number of perfect square numbers 
        // which sum to i. Note that cntPerfectSquares[0] is 0.
        vector<int> cntPerfectSquares(n + 1, INT_MAX);
        cntPerfectSquares[0] = 0;
        for (int i = 1; i <= n; i++)
        {
            // For each i, it must be the sum of some number (i - j*j) and 
            // a perfect square number (j*j).
            for (int j = 1; j*j <= i; j++)
            {
                cntPerfectSquares[i] = 
                    min(cntPerfectSquares[i], cntPerfectSquares[i - j*j] + 1);
            }
        }
        
        return cntPerfectSquares.back();
    }

参考：https://www.cnblogs.com/adamwong/p/10202335.html（有对状态方程的解释）
https://blog.youkuaiyun.com/zy2317878/article/details/80884347
https://leetcode.com/problems/perfect-squares/discuss/71488/Summary-of-4-different-solutions-(BFS-DP-static-DP-and-mathematics)（代码参照）

2.还有一种BFS方法没有看懂