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.

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)"

 

 

 

 

 

http://www.cnblogs.com/grandyang/p/4238577.html

 

开始还担心万一结果是无限不循环小数怎么办，百度之后才发现原来可以写成分数的都是有理数，而有理数要么是有限的，要么是无限循环小数，无限不循环的叫无理数，例如圆周率pi或自然数e等

由于还存在正负情况，处理方式是按正数处理，符号最后在判断，那么我们需要把除数和被除数取绝对值，那么问题就来了：由于整型数INT的取值范围是-2147483648～2147483647，而对-2147483648取绝对值就会超出范围，所以我们需要先转为long long型再取绝对值。那么怎么样找循环呢，肯定是再得到一个数字后要看看之前有没有出现这个数。为了节省搜索时间，我们采用哈希表来存数每个小数位上的数字。还有一个小技巧，由于我们要算出小数每一位，采取的方法是每次把余数乘10，再除以除数，得到的商即为小数的下一位数字。等到新算出来的数字在之前出现过，则在循环开始出加左括号，结束处加右括号。代码如下：

 

 

https://leetcode.com/problems/fraction-to-recurring-decimal/discuss/51106/My-clean-Java-solution

 

 

 

The important thing is to consider all edge cases while thinking this problem through, including: negative integer, possible overflow, etc.

 

Use HashMap to store a remainder and its associated index while doing the division so that whenever a same remainder comes up, we know there is a repeating fractional part.