Leetcode-Fraction to Recurring Decimal

Given two integers representing the numerator and denominator of a fraction, return the fraction in string format.

If the fractional part is repeating, enclose the repeating part in parentheses.

For example,

• Given numerator = 1, denominator = 2, return "0.5".
• Given numerator = 2, denominator = 1, return "2".
• Given numerator = 2, denominator = 3, return "0.(6)"

题目链接 https://leetcode.com/problems/fraction-to-recurring-decimal/

题目的意思其实很简单，给一个除数和被除数，返回商，只不过如果有循环的部分，用括号括起来。

题目本来很简单，一直做除法直到碰到重复出现的余数为止（每次除的时候，把被除数扩大十倍，相当于小数点向后移一位），用一个hashmap来记录余数的出现。

但是，测试用例真心碉堡了。会出现（INT_MIN）/(-1)这样的测试case，因为INT_MIN的绝对值比INT_MAX要大1，所以即使输入的都是int，但是中间结果很可能越界。所以需要使用long long。

有个小技巧，在构造返回值的时候，用ostringstream比直接用string来进行加减，会有效率得多。

//Fraction to Recurring Decimal

string fractionToDecimal(int numerator1, int denominator1){

    long long numerator = (long long)(numerator1);
    long long denominator = (long long)(denominator1);
    ostringstream os;
    if(numerator * denominator < 0)
        os<<"-";
    numerator = numerator > 0 ? numerator : (-numerator);
    denominator = denominator > 0 ? denominator : (-denominator);
    long long zhengshu = numerator / denominator;
    long long  remains = numerator - zhengshu * denominator;
    os<<zhengshu;
    if(remains == 0)
        return os.str();
    int shang;
    vector<int> ans;
    int counter = 0;
    int flag = false;
    map<long long, int> pos; 
    map<long long, int>::iterator it; 
    int position;
    while(remains != 0)
    {
        remains *= 10;

        it = pos.find(remains);
        if(it != pos.end())
        {
            position = it->second;
            flag = true;
            break;
        }
        shang = remains  / denominator;
        ans.push_back(shang);
        pos.insert(make_pair(remains, counter++));
        remains = (remains) - (shang * denominator);
    }
    os<<".";
    for(int i = 0; i < ans.size(); i++)
    {
        if(i == position && flag)
            os<<"(";
        os<<ans[i];
    }
    if(flag)
        os<<")";
    return os.str();
}