原题链接在这里：https://leetcode.com/problems/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.
Example 1:
Input: numerator = 1, denominator = 2 Output: "0.5"
Example 2:
Input: numerator = 2, denominator = 1 Output: "2"
Example 3:
Input: numerator = 2, denominator = 3 Output: "0.(6)"
题解：
加入整数部分后，在reminder 不为0的时候添加小数部分，用HashMap来维护reminder 和 对应的小数长度. 若是出现无限循环可以找到加左括号的起始点.
AC Java:
1 public class Solution { 2 public String fractionToDecimal(int numerator, int denominator) { 3 if(numerator == 0){ 4 return "0"; 5 } 6 if(denominator == 0){ 7 return ""; 8 } 9 10 StringBuilder res = new StringBuilder(); 11 long num = Math.abs((long)numerator); 12 long deno = Math.abs((long)denominator); 13 14 //如果是负数 15 if((numerator < 0) ^ (denominator < 0)){ 16 res.append("-"); 17 } 18 19 res.append(num/deno); 20 long reminder = num%deno; 21 if(reminder == 0){ 22 return res.toString(); 23 } 24 25 //有小数 26 res.append("."); 27 //HashMap中计入reminder 和 对应的小数部分长度，用来找到循环开始的位置 28 HashMap<Long, Integer> hm = new HashMap<Long, Integer>(); 29 StringBuilder desRes = new StringBuilder(); 30 while(reminder != 0){ 31 if(hm.containsKey(reminder)){ //出现了循环，找到循环开始的index添加左括号 32 int beginIndex = hm.get(reminder); 33 desRes.insert(beginIndex, "("); 34 desRes.append(")"); 35 res.append(desRes); 36 return res.toString(); 37 } 38 hm.put(reminder, desRes.length()); 39 desRes.append(reminder*10/deno); 40 reminder = (reminder*10)%deno; 41 } 42 res.append(desRes); 43 return res.toString(); 44 } 45 }