Here, We see Fraction to Recurring Decimal LeetCode Solution. This Leetcode problem is done in many programming languages like C++, Java, JavaScript, Python, etc. with different approaches.

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Fraction to Recurring Decimal LeetCode Solution

Problem Statement

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.

If multiple answers are possible, return any of them.

It is guaranteed that the length of the answer string is less than 10⁴ for all the given inputs.

Example 1:
Input: numerator = 1, denominator = 2
Output: “0.5”

Example 2:
Input: numerator = 2, denominator = 1
Output: “2”

Example 3:
Input: numerator = 4, denominator = 333
Output: “0.(012)”

Fraction to Recurring Decimal Leetcode Solution C++

class Solution {
public:
    string fractionToDecimal(int numerator, int denominator) {
        if(!numerator) return "0";
        string ans = "";
        if (numerator > 0 ^ denominator > 0) ans += '-';
        long num = labs(numerator), den = labs(denominator);
        long q = num / den;
        long r = num % den;
        ans += to_string(q);
        if(r == 0) return ans;
        ans += '.';
        unordered_map<long, int> mp;
        while(r != 0){
            if(mp.find(r) != mp.end()){
                int pos = mp[r];
                ans.insert(pos, "(");
                ans += ')';
                break;
            }
            else{
                mp[r] = ans.length();
                r *= 10;
                q = r / den;
                r = r % den;
                ans += to_string(q);
            }
        }
        return ans;
    }
};Code language: PHP (php)

Fraction to Recurring Decimal Leetcode Solution Java

class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        boolean isNegative = (numerator < 0 && denominator > 0 || numerator > 0 && denominator < 0) ? true : false;
        long numeratorL = Math.abs((long) numerator);
        long denominatorL = Math.abs((long) denominator);
        Map<Long, Integer> previousRemains = new HashMap<Long, Integer>();
        StringBuilder sb = new StringBuilder();
        long quotian = numeratorL / denominatorL;
        sb.append(quotian);
        numeratorL %= denominatorL;
        if (numeratorL != 0) {
            sb.append(".");
        }
        int quotianIndex = 0;
        while (numeratorL != 0) {
            numeratorL *= 10;
            quotian = Math.abs(numeratorL / denominatorL);
            if (!previousRemains.containsKey(numeratorL)) {
                sb.append(quotian);
                previousRemains.put(numeratorL, quotianIndex++);
            } else {
                int firstIndex = 1 + previousRemains.get(numeratorL) + sb.indexOf(".");
                sb.insert(firstIndex, '(');
                sb.append(")");
                break;
            }
            numeratorL %= denominatorL;
        }
        if (isNegative) {
            sb.insert(0, "-");
        }
        return sb.toString();
    }
}Code language: JavaScript (javascript)

Fraction to Recurring Decimal Leetcode Solution JavaScript

var fractionToDecimal = function(numerator, denominator) {
    if(!numerator) return '0';
    let str = '';
    if(Math.sign(numerator) !== Math.sign(denominator)) str += '-';
    const numer = Math.abs(numerator)
    const denom = Math.abs(denominator)
    str += Math.floor(numer/denom);
    let rem = numer%denom;
    if(!rem) return str;
    str += '.'
    const map = new Map();
    while(rem !== 0) {
        map.set(rem, str.length);
        rem *= 10;
        str += Math.floor(rem/denom);
        rem %= denom
        if(map.has(rem)) {
            const idx = map.get(rem);
            return str.slice(0, idx) + `(${str.slice(idx)})`; 
        }
    }
    return str;
};Code language: JavaScript (javascript)

Fraction to Recurring Decimal Leetcode Solution Python

class Solution(object):
    def fractionToDecimal(self, numerator, denominator):
        n, remainder = divmod(abs(numerator), abs(denominator))
        sign = '-' if numerator*denominator < 0 else ''
        result = [sign+str(n), '.']
        stack = []
        while remainder not in stack:
            stack.append(remainder)
            n, remainder = divmod(remainder*10, abs(denominator))
            result.append(str(n))
        idx = stack.index(remainder)
        result.insert(idx+2, '(')
        result.append(')')
        return ''.join(result).replace('(0)', '').rstrip('.')Code language: HTML, XML (xml)