LC-592 分数加减

主要问题：

输入一个分数加减的字符串，进行计算并输出分数

 

主要思路：

对字符串分割出每一个分数（包含符号），然后用两个vector来分别存分子和分母。

解析字符串后，对所有分母找他们的最小公倍数，然后对分子进行相应的加倍后加减，算出总和。

最后计算分子总和与分母最小公倍数的最大公约数，进行化简，输出答案。

 

代码：

class Solution {
public:
    int gcd(int a, int b) {
        if(b) while((a %= b) && (b %= a));
        return a + b;
    }
    
    int lcm(int a, int b){
        return (a * b) / gcd(a, b);
    }
    
    string fractionAddition(string expression) {
        vector<int> numerator, denominator;
        bool flag = false;
        while (!expression.empty()) {
            int index;
            for (index = 0; index < expression.length(); index++) {
                if (index != 0 && (expression[index] == '+' || expression[index] == '-')) {
                    break;
                }
            }
            string fraction = expression.substr(0, index);
            expression.erase(0, index);
            int split = fraction.find('/');
            string num = fraction.substr(0, split);
            string deno = fraction.substr(split + 1);
            numerator.push_back(stoi(num));
            denominator.push_back(stoi(deno));
        }
        int lcm_num = denominator[0];
        for (int i = 1; i < denominator.size(); i++) {
            int temp = lcm(lcm_num, denominator[i]);
            if (temp > lcm_num) lcm_num = temp;
        }
        int sum = 0;
        for (int i = 0; i < numerator.size(); i++) {
            numerator[i] = numerator[i] * (lcm_num / denominator[i]);
            sum += numerator[i];
        }
        int gcd_num = gcd(abs(sum), lcm_num);
        string answer = to_string(sum / gcd_num) + "/" + to_string(lcm_num / gcd_num);
        return answer;
    }
};