[LeetCode] 167. Fraction to Recurring Decimal 分数转循环小数

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)"

给2个整数分别作分子和分母，返回分数的字符串形式。

Java:

public class Solution {
    public String fractionToDecimal(int numerator, int denominator) {
        if (numerator == 0) {
            return "0";
        }
        StringBuilder res = new StringBuilder();
        // "+" or "-"
        res.append(((numerator > 0) ^ (denominator > 0)) ? "-" : "");
        long num = Math.abs((long)numerator);
        long den = Math.abs((long)denominator);

        // integral part
        res.append(num / den);
        num %= den;
        if (num == 0) {
            return res.toString();
        }

        // fractional part
        res.append(".");
        HashMap<Long, Integer> map = new HashMap<Long, Integer>();
        map.put(num, res.length());
        while (num != 0) {
            num *= 10;
            res.append(num / den);
            num %= den;
            if (map.containsKey(num)) {
                int index = map.get(num);
                res.insert(index, "(");
                res.append(")");
                break;
            }
            else {
                map.put(num, res.length());
            }
        }
        return res.toString();
    }
}

Python:

class Solution(object):
    def fractionToDecimal(self, numerator, denominator):
        """
        :type numerator: int
        :type denominator: int
        :rtype: str
        """
        result = ""
        if (numerator > 0 and denominator < 0) or (numerator < 0 and denominator > 0):
            result = "-"

        dvd, dvs = abs(numerator), abs(denominator)
        result += str(dvd / dvs)
        dvd %= dvs

        if dvd > 0:
            result += "."

        lookup = {}
        while dvd and dvd not in lookup:
            lookup[dvd] = len(result)
            dvd *= 10
            result += str(dvd / dvs)
            dvd %= dvs

        if dvd in lookup:
            result = result[:lookup[dvd]] + "(" + result[lookup[dvd]:] + ")"

        return result

C++:

// upgraded parameter types
string fractionToDecimal(int64_t n, int64_t d) {
    // zero numerator
    if (n == 0) return "0";

    string res;
    // determine the sign
    if (n < 0 ^ d < 0) res += ‘-‘;

    // remove sign of operands
    n = abs(n), d = abs(d);

    // append integral part
    res += to_string(n / d);

    // in case no fractional part
    if (n % d == 0) return res;

    res += ‘.‘;

    unordered_map<int, int> map;

    // simulate the division process
    for (int64_t r = n % d; r; r %= d) {

        // meet a known remainder
        // so we reach the end of the repeating part
        if (map.count(r) > 0) {
            res.insert(map[r], 1, ‘(‘);
            res += ‘)‘;
            break;
        }

        // the remainder is first seen
        // remember the current position for it
        map[r] = res.size();

        r *= 10;

        // append the quotient digit
        res += to_string(r / d);
    }

    return res;
}

原文地址：https://www.cnblogs.com/lightwindy/p/9770473.html