Leetcode-166. Fraction to Recurring Decimal

前言：为了后续的实习面试，开始疯狂刷题，非常欢迎志同道合的朋友一起交流。因为时间比较紧张，目前的规划是先过一遍，写出能想到的最优算法，第二遍再考虑最优或者较优的方法。如有错误欢迎指正。博主首发CSDN，mcf171专栏。

博客链接：mcf171的博客

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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)".

Hint:

1. No scary math, just apply elementary math knowledge. Still remember how to perform a long division?
2. Try a long division on 4/9, the repeating part is obvious. Now try 4/333. Do you see a pattern?
3. Be wary of edge cases! List out as many test cases as you can think of and test your code thoroughly.

这个题很难，参考了论坛的解法才做出来的。原文链接：点击打开链接

主要有很多注意事项，其中最关键的是如何对循环数进行处理，同时还有就是注意负数变成正数会越界的问题。Your runtime beats 50.17% of java submissions.

public class Solution {    public String fractionToDecimal(int numerator, int denominator) {        if(numerator == 0) return "0";        StringBuffer sb = new StringBuffer("");        sb.append((numerator > 0) ^ (denominator > 0)?"-":"" );        long num = Math.abs((long)numerator);        long den = Math.abs((long)denominator);        sb.append(num/den);        num %= den;        if(num == 0) return sb.toString();        sb.append(".");        Map<Long,Integer> str2index = new HashMap<Long,Integer>();        str2index.put(num,sb.length());        while(num != 0){            num *= 10;            sb.append(num/den);            num %= den;            if(str2index.containsKey(num)){                int index = str2index.get(num);                sb.insert(index,"(");                sb.append(")");                break;            }else{                str2index.put(num,sb.length());            }        }        return sb.toString();    }}