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。