Leetcode Fraction to Recurring Decimal

最新推荐文章于 2020-08-12 17:50:08 发布

原创最新推荐文章于 2020-08-12 17:50:08 发布 · 144 阅读

0 ·

CC 4.0 BY-SA版权

leetcode 同时被 2 个专栏收录

146 篇文章

订阅专栏

leetcode-math

8 篇文章

订阅专栏

本文提供了一个方法，用于将给定的整数对（分子和分母）转换为小数字符串形式，并正确标识循环节。通过使用哈希映射来跟踪已出现的余数及其位置，该方法能够高效地识别并表示循环部分。

摘要生成于 C知道，由 DeepSeek-R1 满血版支持，前往体验 >

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)”.

math类的问题肯定要考虑的就是符号和overflow. 这里overflow直接用long解决
计算循环节的方法：如果出现已经出现过的余数，找到对应的index，到当前index之间的就是循环节
[code]

public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if(numerator==0)return new String("0");
        if(denominator==0)return new String();
        if(denominator==1)return String.valueOf(numerator);
        HashMap<Long, Integer> map=new HashMap<Long, Integer>();
        StringBuffer sb=new StringBuffer();
        int sign= (numerator>0 && denominator>0) || (numerator<0 && denominator<0) ? 1:-1;
        long num=Math.abs((long)numerator), den=Math.abs((long)denominator);
        int point=-1;
        map.put(num,-1);
        while(num!=0)
        {
            if(num<den)
            {
                sb.append('0');
            }
            else
            {
                sb.append(String.valueOf((long)(num/den)));
                num=num%den;
            }
            num*=10;
            if(point==-1)point=sb.length();
            if(map.containsKey(num))
            {
                sb.insert(map.get(num),"(");
                sb.insert(point,'.');
                sb.insert(sb.length(),')');
                if(sign==-1)sb.insert(0,'-');
                return sb.toString();
            }
            else
            {
                map.put(num, sb.length());
            }

        }
        if(point<sb.length())sb.insert(point,'.');
        if(sign==-1)sb.insert(0,'-');
        return sb.toString();
    }
}