leetcode 166. 分数到小数(Fraction to Recurring Decimal)

题目描述：

给定两个整数，分别表示分数的分子 numerator 和分母 denominator，以字符串形式返回小数。

如果小数部分为循环小数，则将循环的部分括在括号内。

示例 1:

输入: numerator = 1, denominator = 2
输出: "0.5"

示例 2:

输入: numerator = 2, denominator = 1
输出: "2"

示例 3:

输入: numerator = 2, denominator = 3
输出: "0.(6)"

---

解法：

class Solution {
public:
    string itoa(long long num){
        string res = "";
        if(num == 0){
            return "0";
        }else{
            while(num != 0){
                res = char(num%10 + '0') + res;
                num /= 10;
            }
            return res;
        }
    }
    
    string fractionToDecimal(int num, int den) {
        long long numerator = num;
        long long denominator = den;
        if(numerator == 0){
            return "0";
        }
        bool neg = false;
        if(numerator < 0){
            neg = !neg;
            numerator = -numerator;
        }
        if(denominator < 0){
            denominator = -denominator;
            neg = !neg;
        }
        
        unordered_map<int, int> mp;
        long long left = numerator/denominator;
        // cout<<left<<endl;
        numerator %= denominator;
        vector<int> lst;
        int idx = 0;
        while(numerator != 0 && mp.find(numerator) == mp.end()){
            // cout<<numerator<<endl;
            mp[numerator] = idx++;
            numerator *= 10;
            int digit = numerator/denominator;
            numerator %= denominator;
            // cout<<digit<<", "<<numerator<<", "<<denominator<<", "<<idx<<endl;
            lst.push_back(digit);
        }
        
        string res = itoa(left);
        if(neg){
            res = "-" + res;
        }
        
        if(mp.empty()){
            return res;
        }else{
            res += '.';
            if(numerator == 0){
                for(int digit : lst){
                    res += char('0' + digit);
                }
                return res;
            }else{
                int idx = mp[numerator];
                int sz = lst.size();
                for(int i = 0; i < idx; i++){
                    res += char('0' + lst[i]);
                }
                res += '(';
                for(int i = idx; i < sz; i++){
                    res += char('0' + lst[i]);
                }
                res += ')';
                return res;
            }
        }
    }
};