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.
思路：除了2和3以外，最终的结果是2和3的乘积，3多2少。
代码如下：
1 public class Solution { 2 public int integerBreak(int n) { 3 4 if(n==2) 5 return 1; 6 if(n==3) 7 return 2; 8 9 int result=1; 10 if(n%3==2) 11 {result=2; 12 n=n-2; 13 } 14 else if(n%3==1) 15 { 16 result=4; 17 n=n-4; 18 } 19 int count=n/3; 20 while(count>0) 21 { 22 result=result*3; 23 count--; 24 } 25 return result; 26 } 27 }