這種題都要設置一個符號位的變量

224. Basic Calculator

設置數值和符號兩個變量，遇到左括號將數值和符號加進棧中

class Solution {
public:
    int calculate(string s) {
        stack<int> st;
        int res = 0,flag = 1;
        for(int i = 0;i < s.size();i++){
            if(s[i] >= ‘0‘ && s[i] <= ‘9‘){
                
 
int num = 0;
                while(i < s.size() && s[i] >= ‘0‘ && s[i] <= ‘9‘){
                    num = num*10 + flag * (s[i] - ‘0‘);
                    i++;
                }
                res += num;
                i--;
            }
            else if(s[i] == ‘
 
+‘)
                flag = 1;
            else if(s[i] == ‘-‘)
                flag = -1;
            else if(s[i] == ‘(‘){
                st.push(res);
                st.push(flag);
                res = 0,flag = 1;
            }
            else if(s[i] == ‘)‘){
                res *= st.top();
                st.pop();
                res 
 
+= st.top();
                st.pop();
            }
        }
        return res;
    }
};

227. Basic Calculator II

乘除法有優先級，這個時候需要將這些數值彈出

class Solution {
public:
    int calculate(string s) {
        long res = 0,num = 0;
        char op = ‘+‘;
        stack<int> st;
        for(int i = 0;i < s.size();i++){
            if(s[i] >= ‘0‘ && s[i] <= ‘9‘)
                num = num * 10 + s[i] - ‘0‘;
            if(s[i] == ‘+‘ || s[i] == ‘-‘ || s[i] == ‘*‘ || s[i] == ‘/‘ || i == s.size() - 1){
                if(op == ‘+‘)
                    st.push(num);
                if(op == ‘-‘)
                    st.push(-num);
                if(op == ‘*‘ || op == ‘/‘){
                    int tmp = op == ‘*‘ ? st.top() * num : st.top() / num;
                    st.pop();
                    st.push(tmp);
                }
                op = s[i]; 
                num = 0;
            }
        }
        while(!st.empty()){
            res += st.top();
            st.pop();
        }
        return res;
    }
};

leetcode 224. Basic Calculator 、227. Basic Calculator II