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 and not larger than 58.

class Solution {public:    int integerBreak(int n) {        if(n < 4) return n-1;        int res = 1;        while(n > 2){//看n包含多少个3,把他们相乘,直到n<=2            res *= 3;            n -= 3;        }        if(n == 0) return res;//n可以整除3，res就是各个3相乘        if(n == 1) return (res / 3 ) * 4;//除3余1，把其中的一个3加1变为4再相乘        if(n == 2) return res * 2;//除3余2,则可直接把2与res相乘    }};