[leetcode] 166. 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)".

这道题是分数转换为小数，题目难度为Medium。

要将分数转换为小数，可以从小数的计算过程着手。以1/8为例，分为以下几步：

• 商为0，余数为1，记录当前的商，开始进入小数部分，至此小数为“0.”

• 拿余数1乘10作为新的分子继续计算，商为1，余数为2，至此小数为“0.1”

• 拿余数2乘10作为新的分子继续计算，商为2，余数为4，至此小数为“0.12”

• 拿余数4乘10作为新的分子继续计算，商为5，余数为0，计算结束，最终小数为“0.125”

题目的关键部分在于判断分数是否是无限循环小数，如果计算过程中当前余数和之前的余数相同，说明小数开始循环，就可以结束计算了。要判断余数是否重复出现过，自然会想到用Hash Table。另外题目还需要注意两点：分子和分母有可能是负数，如果符号不同结果中要有负号；分数是无限循环小数时要在结果中加入括号，因此需要在Hash Table中记录下余数出现的位置，重复出现时在该位置之前插入左括号。运算结束的条件是余数为0或者余数出现重复。具体代码：

class Solution {public:    string fractionToDecimal(int numerator, int denominator) {        string rst;        unordered_map<long long, int> hash;        int pos;        long long quotient, remainder, curNum, curDen;        if((numerator < 0 && denominator > 0) || (numerator > 0 && denominator < 0)) rst += "-";        if(numerator < 0) curNum = abs((long long)numerator);        else curNum = numerator;        if(denominator < 0) curDen = abs((long long)denominator);        else curDen = denominator;        quotient = curNum / curDen;        remainder = curNum % curDen;                rst += to_string(quotient);        if(remainder) rst += ".";                pos = rst.size();        hash[remainder] = pos;                while(remainder) {            curNum = remainder * 10;            remainder = curNum % curDen;            quotient = curNum / curDen;            rst += to_string(quotient);            if(remainder == 0) {                break;            }            else {                auto it = hash.find(remainder);                if(it != hash.end()) {                    rst.insert(it->second, "(");                    rst += ")";                    break;                }                else {pos++;                    hash[remainder] = pos;                }            }        }                return rst;    }};