Fraction to Recurring Decimal leetcode

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)".

Credits:
Special thanks to @Shangrila for adding this problem and creating all test cases.

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这个题的思路简单，
1.根据两个数得到结果正负，如负数，添加"-"
2.计算小数点前部分，添加到字符串
3.循环计算小数点后部分，并利用hashmap记录每次的除数，如果在hashmap中存在过，说明存在循环小数部分
4.将循环部分括起来

第一次写出的答案如下：

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。