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.

Hint:

1. There is a simple O(n) solution to this problem.
2. You may check the breaking results of n ranging from 7 to 10 to discover the regularities.

方法一：列舉n種拆分方法。

public class Solution {
    public int integerBreak(int n) {
        int max = 1;
        for(int i=2; i<n; i++) {
            int product = 1;
            int m=n/i, remain = n-n/i*i;
            for(int j=0; j<i; j++) {
                if (remain == 0) {
                    product *= m;
                } else {
                    product *= m + 1;
                    remain --;
                }
            }
            if (product < max) break;
            max = product;
        }
        return max;
    }
}
 

方法二：深度優先搜尋。

public class Solution {
    private int max = 0;
    private void find(int from, int n, int product, int step) {
        if (n==0) {
            if (step>1) max = Math.max(max, product);
            return;
        }
        for(int i=from; i<=3 && i<=n; i++) {
            find(i, n-i, product*i, step+1);
        }
    }
    public int integerBreak(int n) {
        find(1, n, 1, 0);
        return max;
    }
}
 

public class Solution {
    public int integerBreak(int n) {
        if (n<2) return 0;
        if (n==2) return 1;
        if (n==3) return 2;
        int p = 1;
        while (n>4) {
            p *= 3;
            n -= 3;
        }
        p *= n;
        return p;
    }
}