leetcode 343. Integer Break

原题：

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.

代码如下：

int integerBreak(int n) {    int sum=1;    int flag=0;    if(n==2)    {        sum=sum;        flag=1;    }    if(n==3)    {        sum=sum*2;        flag=1;    }    if(n==4)    {        sum=sum*4;        flag=1;    }    if(n==5)    {        sum=sum*6;        flag=1;    }    if(flag==1)        n=n-6;    while(n>0)    {        if(n==1)        {            sum=sum/3*4;        }        if(n==2)        {            sum=sum*2;        }        if(n==3)        {            sum=sum*3;        }        if(n==4)        {            sum=sum*4;        }        if(n==5)        {            sum=sum*6;        }        if(n>=6)        {            sum=sum*9;        }        n=n-6;    }    return sum;}

其实就是找规律，发现6的循环是最大的。然后就是各种具体情况进行分析了。