150. Evaluate Reverse Polish Notation

Evaluate the value of an arithmetic expression in Reverse Polish Notation.

Valid operators are +, -, *, /. Each operand may be an integer or another expression.

Note:

• Division between two integers should truncate toward zero.

• The given RPN expression is always valid. That means the expression would always evaluate to a result and there won't be any divide by zero operation.

Example 1:

Input: ["2", "1", "+", "3", "*"]
Output: 9
Explanation: ((2 + 1) * 3) = 9

Example 2:

Input: ["4", "13", "5", "/", "+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6

Example 3:

Input: ["10", "6", "9", "3", "+", "-11", "*", "/", "*", "17", "+", "5", "+"]
Output: 22
Explanation: 
  ((10 * (6 / ((9 + 3) * -11))) + 17) + 5
= ((10 * (6 / (12 * -11))) + 17) + 5
= ((10 * (6 / -132)) + 17) + 5
= ((10 * 0) + 17) + 5
= (0 + 17) + 5
= 17 + 5
= 22

// Iteration
int evalRPN(vector<string>& tokens) { // time: O(n); space: O(n)
    if (tokens.empty()) return 0;
    int n = tokens.size();
    stack<int> st;
    for (string& token : tokens) {
        if (token != "+" && token != "-" && token != "*" && token != "/") {
            st.push(stoi(token));
        } else {
            int num2 = st.top(); st.pop();
            int num1 = st.top(); st.pop();
            if (token == "+") st.push(num1 + num2);
            else if (token == "-") st.push(num1 - num2);
            else if (token == "*") st.push(num1 * num2);
            else if (token == "/") st.push(num1 / num2);
        }
    }
    return st.top();
}

Reverse Polish Notation最後一定是一個operator，然後operand2和operand1，所以可以從後往前進行DFS。

// Recursion
int helper(vector<string>& tokens, int& idx) {
    string str = tokens[idx];
    if (str != "+" && str != "-" && str != "*" && str != "/") return stoi(str);
    int num2 = helper(tokens, --idx), num1 = helper(tokens, --idx);
    if (str == "+") return num1 + num2;
    else if (str == "-") return num1 - num2;
    else if (str == "*") return num1 * num2;
    else return num1 / num2;
}
int evalRPN(vector<string>& tokens) { // time: O(n); space: O(n)
    int idx = tokens.size() - 1;
    return helper(tokens, idx);
}