LeetCode----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.

代码：

class Solution {public:    int integerBreak(int N)     {        int left = 0;        int product = 1;        if(N<2)        {            return -1;        }        else        {            switch(N)            {            case 2:                return 1;                break;            case 3:                return 2;                break;            case 4:                return 4;                break;            default:                while(N>4)                {                    N-=3;                    product*=3;                }                left = product*N;                return left;                break;            }        }       }};

思路：

当N大于4以后，每次分出3后所得到的乘积是最大的。