[leetcode 166] Fraction to Recurring Decimal

最新推荐文章于 2025-12-28 17:52:30 发布

原创最新推荐文章于 2025-12-28 17:52:30 发布 · 414 阅读

0 ·

CC 4.0 BY-SA版权

文章标签：

#leetcode #Hash Table #Math

[LeetCode 刷题] 专栏收录该内容

158 篇文章

订阅专栏

本文介绍了一个算法，该算法接收两个整数作为输入，分别代表一个分数的分子和分母，并返回该分数的小数形式。如果小数部分为循环节，则会使用括号标记循环部分。

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)".

Credits:

Special thanks to @Shangrila for adding this problem and creating all test cases.

class Solution {
public:
    string fractionToDecimal(int numerator, int denominator) {
        long long a = numerator;
        long long b = denominator;
        if (b == 0) {
            return "";
        }
        if (a == 0) {
            return "0";
        }
        string res = "";
        if ((numerator^denominator)>>31) {
            res += '-';
            a = abs(a);
            b = abs(b);
        }
        long long q = a/b;
        if (q == 0) {
            res += "0";
        } else {
            res += to_string(q);
            if (a%b == 0) {
                return res;
            }
            a -= q*b;
        }
        res += ".";
        unordered_map<long long, int> mapping;
        while (a) {
            if (mapping.find(a) != mapping.end()) {
                res.insert(mapping[a], 1, '(');
                res += ")";
                break;
            }
            mapping[a] = res.size();
            a = a * 10;
            res += to_string(a/b);
            a = a%b;
        }
        return res;
    }
};