Fraction to Recurring Decimal

Fraction to Recurring Decimal

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.

思路：

　　模拟除法

　　判断循环节的时候用hashtable

我的代码：

import java.util.Hashtable;
public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if(denominator == 0 || numerator == 0 )    return "0";
        int cnt = 0;
        if(numerator < 0)
        {
            cnt++;
        }
        if(denominator < 0)
        {
            cnt++;
        }
        long num = Math.abs((long)numerator);
        long den = Math.abs((long)denominator);
        StringBuffer left = new StringBuffer();
        long remain = num%den;
        num /= den;
        left.append(num);
        if(remain == 0) 
        {
            return cnt == 1 ? "-"+left.toString() : left.toString();
        }
        StringBuffer right = new StringBuffer();
        Hashtable<Long,Integer> ht = new Hashtable<Long,Integer>();
        num = remain;
        ht.put(num, 0);
        int index = 1;
        while(num != 0)
        {
            num *= 10; 
            remain = num%den;
            num /= den;
            right.append(num);
            if(ht.containsKey(remain))
            {
                right.insert(ht.get(remain),"(");
                right.append(')');
                break;
            }
            ht.put(remain,index);
            num = remain;
            index++;
        }
        String rst = left.toString() + "." + right.toString();
        return cnt == 1 ? "-"+rst : rst;
    }
}

他人代码：

public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) {
            return "0";
        }
        StringBuilder res = new StringBuilder();
        // "+" or "-"
        res.append(((numerator > 0) ^ (denominator > 0)) ? "-" : "");
        long num = Math.abs((long)numerator);
        long den = Math.abs((long)denominator);

        // integral part
        res.append(num / den);
        num %= den;
        if (num == 0) {
            return res.toString();
        }

        // fractional part
        res.append(".");
        HashMap<Long, Integer> map = new HashMap<Long, Integer>();
        map.put(num, res.length());
        while (num != 0) {
            num *= 10;
            res.append(num / den);
            num %= den;
            if (map.containsKey(num)) {
                int index = map.get(num);
                res.insert(index, "(");
                res.append(")");
                break;
            }
            else {
                map.put(num, res.length());
            }
        }
        return res.toString();
    }
}

学习之处：

• 这道题的思路很简单，看到题的那一刻基本上也知道怎么做了，但是需要考虑的corner case 特别多
  • 分母为0 分子为0
  • 永远记住对于int而言负数转换成正数的时候，需要用long进行存储，因为会溢出啊！！
  • 循环节问题 仔细分析 1/6 和 2/3 这两种情况就可以了
• 判断乘积或者除法的正负号的时候，用这种方法真简洁 (numerator > 0) ^ (denominator > 0)) ? "-" : "