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.
题意&解题思路
将一个数分解为若干个数的和，使得分解后的数乘积最大。 将输入的数分解为2和3即可（4 = 2 + 2 = 2 * 2）。代码如下：
class Solution {
public:
int integerBreak(
int n) {
if(n ==
2)
return 1;
if(n ==
3)
return 2;
long long ans =
1;
while(n){
if(n ==
4 || n ==
2){
ans *= n;
break;
}
ans *=
3;
n -=
3;
}
return ans;
}
};
