Leetcode Study Day 27

Evaluate Reverse Polish Notation

You are given an array of strings tokens that represents an arithmetic expression in a Reverse Polish Notation.

Evaluate the expression. Return an integer that represents the value of the expression.

Note that:

The valid operators are ‘+’, ‘-‘, ‘*’, and ‘/‘.
Each operand may be an integer or another expression.
The division between two integers always truncates toward zero.
There will not be any division by zero.
The input represents a valid arithmetic expression in a reverse polish notation.
The answer and all the intermediate calculations can be represented in a 32-bit integer.

Example 1:

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

Input: tokens = ["4","13","5","/","+"]
Output: 6
Explanation: (4 + (13 / 5)) = 6
Example 3:

Input: tokens = ["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
 

Constraints:

1 <= tokens.length <= 104
tokens[i] is either an operator: "+", "-", "*", or "/", or an integer in the range [-200, 200].

My solution with stack

Based on the definition of RPN, I found that we can use a stack to store the values follow the order of the input vector, once we find an operator, we pop the top two values from the stack and calculate the result, then push the result back to the stack. After the iteration, we return the top of the stack.

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        stack<int> RPNstack;
        size_t n = tokens.size();
        for (int i = 0; i < n; i++){
            if(tokens[i] != "+" && tokens[i] != "-" && tokens[i] != "*" && tokens[i] != "/"){
                int num = stoi(tokens[i]);
                RPNstack.push(num);
            }
            else if(tokens[i] == "+"){
                int temp1 = RPNstack.top();
                RPNstack.pop();
                int temp2 = RPNstack.top();
                RPNstack.pop();
                int result = temp2 + temp1;
                RPNstack.push(result);
            }
            else if(tokens[i] == "-"){
                int temp1 = RPNstack.top();
                RPNstack.pop();
                int temp2 = RPNstack.top();
                RPNstack.pop();
                int result = temp2 - temp1;
                RPNstack.push(result);
            }
            else if(tokens[i] == "*"){
                int temp1 = RPNstack.top();
                RPNstack.pop();
                int temp2 = RPNstack.top();
                RPNstack.pop();
                int result = temp2 * temp1;
                RPNstack.push(result);
            }
            else if(tokens[i] == "/"){
                int temp1 = RPNstack.top();
                RPNstack.pop();
                int temp2 = RPNstack.top();
                RPNstack.pop();
                int result = temp2 / temp1;
                RPNstack.push(result);
            }
        } 
        return RPNstack.top();
    }
};

Solution with unordered_map

I found a solution using unordered_map and stack to solve this problem. One thing I’ve never seen before is the use of unordered_map. I think it is a good way to store the operator and the corresponding function.

When we iterate the vector, if the element is not an operator, we push it into the stack. If it is an operator, we pop the top two elements from the stack and calculate the result by mapping the operator to the responding operation, then push the result back to the stack. After the iteration, we return the top of the stack.

class Solution {
public:
    int evalRPN(vector<string>& tokens) {
        unordered_map<string, function<int (int, int) > > map = {
            { "+" , [] (int a, int b) { return a + b; } },
            { "-" , [] (int a, int b) { return a - b; } },
            { "*" , [] (int a, int b) { return a * b; } },
            { "/" , [] (int a, int b) { return a / b; } }
        };
        std::stack<int> stack;
        for (string& s : tokens) {
            if (!map.count(s)) {
                stack.push(stoi(s));
            } else {
                int op1 = stack.top();
                stack.pop();
                int op2 = stack.top();
                stack.pop();
                stack.push(map[s](op2, op1));
            }
        }
        return stack.top();
    }
};