Leetcode: Fraction to Recurring Decimal

题目：
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)".

思路：
用一个map记录每一个余数，当出现重复的余数时，那么将会进入循环，两个重复余数之间的部分就是循环体。

C++代码：

class Solution
{
public:
    string fractionToDecimal(int numerator, int denominator)
    {
        if (numerator == 0) return "0";
        if (denominator == 0) return "";

        string result = "";
        //如果其中有一个数为负，反正结果添加负号
        if ((numerator < 0) ^ (denominator < 0)) result += "-";

        //给分子和分母取绝对值
        long long lnumerator= numerator;
        long long ldenominator = denominator;
        lnumerator = abs(lnumerator);
        ldenominator = abs(ldenominator);

        long long quotient = lnumerator / ldenominator;
        result += to_string(quotient);
        //给余数乘以10是为了后续计算余数的结果为整数
        long long remainder = lnumerator % ldenominator * 10;
        //如果余数为0，直接返回结果
        if (remainder == 0) return result;

        result += ".";
        unordered_map<long long, int> dictionary;//用于存储余数及其余数的下标
        while (remainder != 0)
        {
            //如果dictionary中存在这个余数
            if (dictionary.find(remainder) != dictionary.end())
            {
                int position = dictionary[remainder];
                string front = result.substr(0, position);
                string back = result.substr(position, result.length());
                return front + "(" + back + ")";
            }
            dictionary.insert({remainder, result.length()});
            quotient = remainder / ldenominator;//用短除法计算商
            result += to_string(quotient);
            remainder = remainder % ldenominator * 10;//得到余数并乘以10
        }
        return result;
    }
};

Java代码（因为这道题没有提供C#代码的测试，所以这里使用Java，两个语言的语法差异不是太大）：

public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) return "0";
        if (denominator == 0) return "";

        StringBuilder result = new StringBuilder();
        //如果numerator和denominator中有一个为负数，则结果添加负号
        if ((numerator > 0) ^ (denominator > 0)) result.append("-");

        //给numerator和denominator取绝对值
        long lnumberator = numerator;
        long ldenominator = denominator;
        lnumberator = Math.abs(lnumberator);
        ldenominator = Math.abs(ldenominator);

        //计算整数部分
        long quotient = lnumberator /ldenominator;
        result.append(quotient);
        //余数乘以10是为了让后面余数部分的计算
        long remainder = lnumberator % ldenominator * 10;
        //如果余数为0直接返回result结果
        if (remainder == 0) return result.toString();

        result.append(‘.‘);
        Map<Long, Integer> dictionary = new HashMap<Long, Integer>();
        while (remainder != 0) {
            //如果dictionary中存在remainder的值，则说明小数部分出现循环
            if (dictionary.containsKey(remainder)) {
                int position = dictionary.get(remainder);
                String front = result.substring(0, position);
                String back = result.substring(position, result.length());
                return front + ‘(‘ + back + ‘)‘;
            }

            //将remainder添加到dictionary中(key为remainder，value为对应的在result中的位置)，继续计算余数
            dictionary.put(remainder, result.length());
            quotient = remainder / ldenominator;
            result.append(quotient);
            remainder = remainder % ldenominator * 10;
        }

        return result.toString();
    }
}

Python代码：

class Solution:
    # @return a string
    def fractionToDecimal(self, numerator, denominator):
        if numerator == 0:
            return ‘0‘
        if denominator == 0:
            return ‘‘

        result = ‘‘
        if (numerator > 0) ^ (denominator > 0):
            result = result + ‘-‘

        pnumerator = abs(numerator)
        pdenominator = abs(denominator)

        quotient = pnumerator / pdenominator
        result = result + str(quotient)
        remainder = pnumerator % pdenominator * 10

        if remainder == 0:
            return result

        result = result + ‘.‘
        map = {}
        while remainder != 0:
            if map.has_key(remainder):
                position = map[remainder]
                front = result[0: position]
                back = result[position: len(result)]
                return front + ‘(‘ + back + ‘)‘

            map[remainder] = len(result)
            quotient = remainder / pdenominator
            result = result + str(quotient)
            remainder = remainder % pdenominator * 10
        return result