Fraction to Recurring Decimal

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)".

    题意： 给定两个整型数，一个代表分子 numerator ，一个代表分母 denominator，以小数的形式返回它们的结果result，当有循环小数时，以括号形式表示。比如5/3=1.666666...以“1.(6)”的形式返回，返回的类型是字符串。

解题要点：

1、注意两数的正负号，保证求得结果的符号正确性。

2、INT_MIN的处理，将 INT_MIN转化为正数会溢出，即-2147483648变成整数会溢出，因此要使用long long int来计算。

      long long int A = abs(numerator), B = abs(denominator);若denominator = -2147483648，则B仍为-2147483648应该改为 long long int A = numerator, B = denominator;         A = abs(A);         B = abs(B);    或者         long long int A = abs((long long int)numerator), B = abs((long long int)denominator);

3、分为整数部分和小数部分，重点在于小数部分的处理，因为小数部分有可能会出现循环。观察下面的除法：

0.16  6 ) 1.00  0   1 0   <-- Remainder=1, mark 1 as seen at position=0.  - 6     40  <-- Remainder=4, mark 4 as seen at position=1.  - 36      4  <-- Remainder=4 was seen before at position=1, so the fractional part which is 16 starts repeating at position=1 => 1(6).

当余数为4时，将出现循环，所以我们可以设置一个哈希表，存储每一次的余数，以及该余数在返回结果result中的下标。每一次得到新的余数，就查询该余数是否已经在哈希表中，是的话说明开始循环了，那么直接在result中该余数对应的位置后面插入‘（’，result末尾加上‘）’，结束运算。如果在哈希表中没找到，则继续正常运运算。

C++代码：

#include <iostream> #include <unordered_map>#include <sstream>#include <string>#include <cmath>using namespace std;  class Solution {public:  string fractionToDecimal(int numerator, int denominator)   {    if(0 == numerator)return "0";    string result;if(numerator < 0 ^ denominator < 0)//判断正负号,异或只有一个为真result = result + "-";/**************************************************************long long int A = abs(numerator), B = abs(denominator);若denominator = -2147483648，则B仍为-2147483648应该改为 long long int A = numerator, B = denominator;         A = abs(A);         B = abs(B);    或者         long long int A = abs((long long int)numerator), B = abs((long long int)denominator);***************************************************************/long long int A = numerator, B = denominator;//防止将INT_MIN转化为正数会溢出A = abs(A);B = abs(B);unordered_map<long long int, long long int> object;//存放余数和标号long long int num1 = 0;//整数部分的位数long long int num2 = -1;//小数部分的位数long long int shang = 0;long long int yushu = 0;while(1){shang = A / B;yushu = A % B;stringstream ss;        ss<<shang;        result = result + ss.str();if(object.count(yushu) != 0)//该余数曾经出现过break;if(-1 == num2)num1 = result.size();//整数部分长度if(0 == yushu)break;else if(-1 == num2)result = result + ".";object.insert(make_pair(yushu, num1 + 2 + num2));//记录余数的位置num2++;    A = yushu * 10;    }if(yushu != 0)//若没有除整{unordered_map<long long int, long long int>::iterator it = object.find(yushu);//找到插入括号的位置result.insert(it->second, "(");result = result + ")";}    return result;  }};int main(int argc, char* argv[])  {  Solution temp;temp.fractionToDecimal(-1, -2147483648);    system("pause");  }

Python代码：

#coding=gbkimport mathclass Solution:    # @return a string    def fractionToDecimal(self, numerator, denominator):        if 0 == numerator:            return "0"        result = ""        dic = {}        if (numerator < 0) ^ (denominator < 0):            result = result + "-"        numerator = abs(numerator)        denominator = abs(denominator)        num1 = 0        num2 = -1        while(1):            shang = numerator / denominator            yushu = numerator % denominator            result = result + str(shang)            if 0 == yushu:                break            if yushu in dic:                result = result[0:dic[yushu]] + "(" + result[dic[yushu]:] + ")"                break            if -1 == num2:                num1 = len(result)                result = result + '.'            dic[yushu] = num1 + 2 + num2            num2 = num2 + 1            numerator = yushu * 10        return resultif __name__ == '__main__':    object = Solution()    result = object.fractionToDecimal(-50, 8)    print result