【编程之美】2.13 子数组的最大乘积

//答案解法
int getMaxProductAnswer(int * a, int alen)
{
    int NumPositive = 0;
    int NumNegtive = 0;
    int NumZero = 0;
    int exceptLoction = 0; //去掉的数字的位置
    int maxProduct = 1;

    for(int i = 0; i < alen; i++)
    {
        if(a[i] == 0)
        {
            NumZero++;
        }
        else if(a[i] > 0)
        {
            NumPositive++;
        }
        else
        {
            NumNegtive++;
        }
    }

    if(NumZero >= 2)
    {
        maxProduct = 0;
    }
    else if(NumZero == 1)
    {
        if(NumNegtive & 0x01 == 1) //奇数个负数
        {
            return 0;
        }
        else //偶数个负数 去掉0
        {
            for(int i = 0; i < alen; i++)
            {
                if(a[i] != 0)
                {
                    maxProduct *= a[i];
                }
            }
        }
    }
    else
    {
        if(NumNegtive & 0x01 == 1) //奇数个负数 去掉负数中绝对值最小的
        {
            int minAbsNegtive = 0;
            int i = 0;
            for(i = 0; i < alen; i++)
            {
                if(a[i] < 0)
                {
                    minAbsNegtive = a[i];
                    exceptLoction = i;
                    break;
                }
            }
            for( ;i < alen; i++)
            {
                if(a[i] < 0 && a[i] > minAbsNegtive)
                {
                    minAbsNegtive = a[i];
                    exceptLoction = i;
                }
            }
            for(i = 0; i < alen; i++)
            {
                if(i != exceptLoction)
                {
                    maxProduct *= a[i];
                }
            }

        }
        else //偶数个负数
        {
            if(NumPositive > 0) //有正数 去掉正数中最小的
            {
                int minPositive = 0;
                int i = 0;
                for(i = 0; i < alen; i++)
                {
                    if(a[i] > 0)
                    {
                        minPositive = a[i];
                        exceptLoction = i;
                        break;
                    }
                }
                for( ;i < alen; i++)
                {
                    if(a[i] > 0 && a[i] < minPositive)
                    {
                        minPositive = a[i];
                        exceptLoction = i;
                    }
                }
                for(i = 0; i < alen; i++)
                {
                    if(i != exceptLoction)
                    {
                        maxProduct *= a[i];
                    }
                }
            }
            else //没有正数 去掉负数绝对值最大的
            {
                int maxAbsNegtive = 0;
                int i = 0;
                for(i = 0; i < alen; i++)
                {
                    if(a[i] < 0)
                    {
                        maxAbsNegtive = a[i];
                        exceptLoction = i;
                        break;
                    }
                }
                for( ;i < alen; i++)
                {
                    if(a[i] < 0 && a[i] < maxAbsNegtive)
                    {
                        maxAbsNegtive = a[i];
                        exceptLoction = i;
                    }
                }
                for(i = 0; i < alen; i++)
                {
                    if(i != exceptLoction)
                    {
                        maxProduct *= a[i];
                    }
                }
            }
        }
    }
    return maxProduct;
}



int main()
{
    int a[10] = {-2,-4,-6,0,-9,-2,-4,-6,-7,-4};
    int maxProduct = getMaxProductAnswer(a, 10);
    return 0;
}