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,
Credits:
Special thanks to @Shangrila for adding this problem and creating all test cases.
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string fractionToDecimal(int numerator, int denominator) { string ret; int a = numerator; int b = denominator; ret = to_string(a / b); a = a % b; if (a) ret.push_back(‘.‘); else return ret; unordered_map<int, int> m; int i = ret.length() - 1; while (a) { i++; if (m.find(a) != m.end()) { ret.insert(ret.begin() + m[a], ‘(‘); ret.push_back(‘)‘); break; } m[a] = i; a *= 10; if (a < b) { ret.push_back(‘0‘); continue; } ret.push_back(‘0‘ + a / b); a = a % b; } return ret; }
提交后发现，有特殊情况没有考虑到。-2147483648 / -1 这种情况会溢出，所以需要使用long类型替换int。
参考修改后代码如下：
string fractionToDecimal(int numerator, int denominator) { if (!numerator) return "0"; string ret; if (numerator < 0 ^ denominator < 0) ret += ‘-‘; long a = numerator < 0 ? (long)numerator * (-1) : (long)numerator; long b = denominator < 0 ? (long)denominator * (-1) : (long)denominator; long c = a / b; ret += to_string(c); a = a % b; if (!a) return ret; ret.push_back(‘.‘); unordered_map<long, long> m; int i = ret.length() - 1; while (a) { i++; if (m.find(a) != m.end()) { ret.insert(ret.begin() + m[a], ‘(‘); ret.push_back(‘)‘); break; } m[a] = i; a *= 10; if (a < b) { ret.push_back(‘0‘); continue; } ret.push_back(‘0‘ + a / b); a = a % b; } return ret; }
虽然在leetcode上提交成功了，但是在vs2015上运行是错误的，
long a = numerator < 0 ? (long)numerator * (-1) : (long)numerator;
这一句如果是numerator是-2147483648，a会是-2147483648，这非常诡异。不知道vs2015编译器对于这个问题是怎么解决的。
Fraction to Recurring Decimal leetcode
原文：http://www.cnblogs.com/sdlwlxf/p/5152236.html