题解 | 逆波兰表达式求值

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> st;
        
        for(string& s : tokens) {
            if(s == "+" || s == "-" || s == "*" || s == "/") {
                int b = st.top(); st.pop();
                int a = st.top(); st.pop();
                
                if(s == "+") st.push(a + b);
                else if(s == "-") st.push(a - b);
                else if(s == "*") st.push(a * b);
                else st.push(a / b);
            } else {
                st.push(stoi(s));
            }
        }
        
        return st.top();
    }
};

import java.util.Stack;

public class Solution {
    public int evalRPN(String[] tokens) {
        Stack<Integer> st = new Stack<>();
        
        for(String s : tokens) {
            if(s.equals("+") || s.equals("-") || 
               s.equals("*") || s.equals("/")) {
                int b = st.pop();
                int a = st.pop();
                
                if(s.equals("+")) st.push(a + b);
                else if(s.equals("-")) st.push(a - b);
                else if(s.equals("*")) st.push(a * b);
                else st.push(a / b);
            } else {
                st.push(Integer.parseInt(s));
            }
        }
        
        return st.peek();
    }
}

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        st = []
        
        for token in tokens:
            if token in ['+', '-', '*', '/']:
                b = st.pop()
                a = st.pop()
                
                if token == '+':
                    st.append(a + b)
                elif token == '-':
                    st.append(a - b)
                elif token == '*':
                    st.append(a * b)
                else:
                    # Python的除法需要特殊处理，确保结果向0取整
                    st.append(int(a / b))
            else:
                st.append(int(token))
        
        return st[-1]