LeetCode之Basic Calculator II

/*用两个辅助vector来进行运算，一个存符号，一个存数字。*/
class Solution {
public:
    int calculate(string s) {
        if(s.empty()) return 0;
        vector<char> op;
        vector<int> num;
        for(size_t i = 0; i < s.size();){
            if(s[i] >= '0' && s[i] <= '9'){//遍历到数字
                int tmp = 0;
                while(i < s.size() && s[i] >= '0' && s[i] <= '9'){
                    tmp = tmp * 10 + (s[i] - '0');
                    ++i;
                }
                if(!op.empty() && (op.back() == '*' || op.back() == '/')){
                    int a = num.back();
                    num.pop_back();
                    num.push_back(calculate(a, tmp, op.back()));
                    op.pop_back();
                }
                else num.push_back(tmp);
            }
            else if(s[i] == '+' || s[i] == '-' || s[i] == '*' || s[i] == '/'){//遇到号，入栈
                op.push_back(s[i]);
                ++i;
            }
            else ++i;
        }
        size_t j = 0;
        for(size_t i = 0; i < op.size(); ++i){
            num[j+1] = calculate(num[j], num[j+1], op[i]);
            ++j;
        }
        return num.back();
    }
    
    int calculate(int a, int b, char op){
        if(op == '+') return a + b;
        else if(op == '-') return a - b;
        else if(op == '*') return a * b;
        else return a / b;
    }
};