给出一个直观的解法，将每个数写成 "+(nom)/(denom)" 的string形式放到deque里，然后再把deque两两计算合并。计算时，要用到gcd来消公约数。写的比较长。

class Solution {
public:
    
    int gcd(int a, int b){
        return b == 0 ? a : gcd(b, a % b);
    }
    
    string calculate(string s1, string s2){

        int idx1 = s1.find_first_of('/'), idx2 = s2.find_first_of('/');
        int nom_1 = stoi(s1.substr(0, idx1)), nom_2 = stoi(s2.substr(0, idx2));
        int denom_1 = stoi(s1.substr(idx1+1)), denom_2 = stoi(s2.substr(idx2+1));
        int res_denom = denom_1 * denom_2; 
        int res_nom = nom_1 * denom_2 + nom_2 * denom_1;
        if(res_nom == 0) return"+0/1";
        
        int common = gcd(abs(res_nom), abs(res_denom));
        res_nom /= common; res_denom /= common;
        
        int sign = (res_nom * res_denom < 0) ? -1 : 1;
        return (sign == 1 ? '+' : '-') + to_string(abs(res_nom)) + '/' + to_string(abs(res_denom));
        
    }

    string fractionAddition(string expression) {
        if(expression.empty()) return "";
        if(isdigit(expression[0])) expression.insert(expression.begin(), '+');
        
        deque<string> dq;
        int start = 0;
        for(int i=1; i<=expression.length(); i++){
            if( i == expression.length() || expression[i] == '+' || expression[i] == '-'){
                string temp = expression.substr(start, i-start);
                dq.push_back(temp);
                start = i;
            }
        }
        while(dq.size() >= 2){
            string s1 = dq.front(); dq.pop_front();
            string s2 = dq.front(); dq.pop_front();
            string res = calculate(s1, s2);
            //cout << s1 << " " << s2 << " " << res << endl;
            dq.push_front(res);
        }
        string res = dq.front();
        return res[0] == '-' ? res : res.substr(1);
    }
};

更简洁的solution参见如下，用regular expression split，再loop数组相加.

https://discuss.leetcode.com/topic/89991/concise-java-solution

其中，String[] fracs = expression.split("(?=[-+])"); 运用了zero-positive look ahead, 表明在+或-的前面split.