Leetcode 343. Integer Break

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.

一道數字遊戲題，從簡單的數字開始動動手就能找到規律。所有正整數可拆分爲1，但乘1不頂用，接下來可用的最基本元素就是2和3，發現用3的積比用2的積大，所以拆分的原則是儘量拆分爲3，湊不夠3了才用2.

方法一：遞歸版本

class Solution {
public:
    int integerBreak(int n) {
        if(n==2) return 1;
        if(n==3) return 2;
        if(n==4) return 4;
        if(n==5) return 6;
        if(n==6) return 9;
        return (3*integerBreak(n-3));
    }
};

方法二：非遞歸版本

class Solution {
public:
    int integerBreak(int n) {
        if (n==2) return 1;
        if (n==3) return 2;          
        if (n%3 == 0) return pow(3,n/3);
        if (n%3 == 2) return pow(3,n/3)*2;
        return pow(3,(n/3)-1)*4;
    }
};