279. Perfect Squares

Given a positive integer n, find the least number of perfect square numbers (for example, 1, 4, 9, 16, …) which sum to n.

For example, given n = 12, return 3 because 12 = 4 + 4 + 4; given n = 13, return 2 because 13 = 4 + 9.

这道题我是用动态规划来做，然后看讨论有人总结了四种方法 (BFS, DP, static DP and mathematics)，参考Summary of 4 different solutions

我的代码：424ms

int numSquares(int n)
{
    static vector<int>dp(n+1,1<<30);
    dp[0]=0;
    for(int i=1;i<=n;i++)
    {
        int m=sqrt(i);
        for(int j=1;j<=m;j++)
        {
            dp[i]=min(dp[i],dp[i-j*j]+1);
        }
    }
    return dp[n];
} 

Static Dynamic Programming: 12ms

class Solution 
{
public:
    int numSquares(int n) 
    {
        if (n <= 0)
        {
            return 0;
        }

        // cntPerfectSquares[i] = the least number of perfect square numbers 
        // which sum to i. Since cntPerfectSquares is a static vector, if 
        // cntPerfectSquares.size() > n, we have already calculated the result 
        // during previous function calls and we can just return the result now.
        static vector<int> cntPerfectSquares({0});

        // While cntPerfectSquares.size() <= n, we need to incrementally 
        // calculate the next result until we get the result for n.
        while (cntPerfectSquares.size() <= n)
        {
            int m = cntPerfectSquares.size();
            int cntSquares = INT_MAX;
            for (int i = 1; i*i <= m; i++)
            {
                cntSquares = min(cntSquares, cntPerfectSquares[m - i*i] + 1);
            }

            cntPerfectSquares.push_back(cntSquares);
        }

        return cntPerfectSquares[n];
    }
};

Mathematical Solution: 4ms

class Solution 
{  
private:  
    int is_square(int n)
    {  
        int sqrt_n = (int)(sqrt(n));  
        return (sqrt_n*sqrt_n == n);  
    }

public:
    // Based on Lagrange's Four Square theorem, there 
    // are only 4 possible results: 1, 2, 3, 4.
    int numSquares(int n) 
    {  
        // If n is a perfect square, return 1.
        if(is_square(n)) 
        {
            return 1;  
        }

        // The result is 4 if and only if n can be written in the 
        // form of 4^k*(8*m + 7). Please refer to 
        // Legendre's three-square theorem.
        while ((n & 3) == 0) // n%4 == 0  
        {
            n >>= 2;  
        }
        if ((n & 7) == 7) // n%8 == 7
        {
            return 4;
        }

        // Check whether 2 is the result.
        int sqrt_n = (int)(sqrt(n)); 
        for(int i = 1; i <= sqrt_n; i++)
        {  
            if (is_square(n - i*i)) 
            {
                return 2;  
            }
        }  

        return 3;  
    }  
};