#leetcode#Fraction to Recurring Decimal

来源：互联网发布：能看伪恋的漫画软件编辑：程序博客网时间：2024/05/17 07:47

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)".

解题思路是首先判断正负，这里用了异或的方法来判断：（numerator > 0）^（denominator > 0） == 1 --> 是负数

然后整数部分很好得到，直接 stringBuilder.append(numerator / denominator) 就可以了，

然后是小数部分的判断，首先 append（'.'），每次计算取余数 remainder，并把remainder作为key，当前 stringbuilder的长度作为value存入hashmap，则当remainder重复出现时我们知道这个除法运算进入了无限循环，加括号跳出循环。

有很多corner case，比如 -1 / -2147483648, 所以用long来防止溢出，

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

0 0