343. Integer Break

题目：
Given a positive integer n, break it into the sum of at least two positive integers and maximize the product of those integers. Return the maximum product you can get.

For example, given n = 2, return 1 (2 = 1 + 1); given n = 10, return 36 (10 = 3 + 3 + 4).

Note: you may assume that n is not less than 2.

解答：
这里我提供了两个办法，一种是我用dp的办法解的，还有一种是参考别人的数学解法：
DP解法：

//State: f[i] is when sum is i the largest product we can get
    //Function: f[i] = max(f[i - j] * j), i >= 2, i - j >= 1
    //Initialize: f[1] = f[2] = 1;
    //Result: f[n]
    public int integerBreak(int n) {
        int[] f = new int[n + 1];
        Arrays.fill(f, 1);
        
        for (int i = 3; i <= n; i++) {
            for (int j = 1; j < i; j++) {
                //注意，f[i]的最大值是由至少两个数的乘积组成的，可是他自己i这个值在比i大的数的乘积中，也能够单独做为一个因子出现，因此要加上跟这种状况的比较，即(i - j) * j
                f[i] = Math.max(f[i], Math.max(f[i - j] * j, (i - j) * j));
            }
        }
        
        return f[n];
    }

数学解法：

//A more mathematic solution
    //We can prove that when N is even, N/2 * N/2 is largest; when N is odd, (N - 1)/2 * (N + 1)/2 is the largest
    //Also we have N/2 * N/2 > N --> N > 4
    //             (N - 1)/2 * (N + 1)/2 > N --> N >= 5
    //So when N > 4, we can do the calculation
    public int integerBreak(int n) {
        if (n == 2) return 1;
        if (n == 3) return 2;
        
        int result = 1;
        while (n > 4) {
            result *= 3;
            n -= 3;
        }
        result *= n;
        return result;
    }