问题

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.

例子

• Given numerator = 1, denominator = 2, return "0.5".

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

分析

模拟除法的运算过程。设numerator = 10， denominator = 6.

1. numerator = 10, denominator = 6, quotient = 1, remander = 4, res = "1.";
2. numerator = remander * 10 = 40, denominator = 6, quotient = 6, remander = 4, res = "1.6";
3. numerator = remander * 10 = 40, denominator = 6, quotient = 6, remander = 4, res = "1.66".

到第三步可以发现，以后每一步remander都是4，quotient都是6，即开始循环。res应该为"1.(6)".

所以我们要模拟上面的运算，直到remander开始重复某一数字，设该数字为R。用一个map来保存remander和它出现的位置。在remander第一次出现R的位置插入'('，在最后的位置插入')'。

除此之外，还有考虑一下几点：

• numerator和denominator异号时，在res前加'-';
• quotient和remander要定义成long long，防止溢出，例如numerator = -2147483648, denominator = -1;
• 求int的绝对值时，有溢出的风险，需要先把int转换成long long.

要点

• 理解小数循环节的含义，能够模拟除法运算：不断用余数乘10除以除数，直到余数为0。当余数开始出现重复数字时，商的小数开始循环；
• 注意数值类型的溢出问题；
• string关于char的构造函数为string(size_t n, char c).

时间复杂度

O(n), n位数字的长度

空间复杂度

O(n)

代码

class Solution {
public:
    string fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) return "0";
        string res;
        if (numerator < 0 ^ denominator < 0)
            res += '-';
        long long numer = abs((long long)numerator);
        long long denom = abs((long long)denominator);
        long long quotient = numer / denom;
        long long remander = numer % denom;
        res += to_string(quotient);
        if (remander == 0) return res;
        res += '.';
        unordered_map<long long, int> map;

        while (remander) {
            quotient = remander * 10 / denom;
            if (map.find(remander) != map.end()) {
                res.insert(map[remander], 1, '(');
                res += ')';
                break;
            }
            map[remander] = res.size();
            res += to_string(quotient);
            remander = remander * 10 % denom;
        }

        return res;
    }
};