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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,
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(); }
Leetcode-Fraction to Recurring Decimal
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原文地址：http://www.cnblogs.com/ouchengeng/p/4324323.html
