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Evaluate Reverse Polish Notation

 문제 설명

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

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

Note that division between two integers should truncate toward zero.

It is guaranteed that the given RPN expression is always valid. That means the expression would always evaluate to a result, and there will not be any division by zero operation.

* 역폴란드 표기법 : https://ko.wikipedia.org/wiki/역폴란드_표기법

 제한 사항

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

 입출력 예

Example 1:

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

더보기

Example 2:

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

더보기

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

 Python 코드

Python code 

class Solution:
    def evalRPN(self, tokens: List[str]) -> int:
        stack = []
        for t in tokens:
            if t not in {"+", "-", "*", "/"}:
                stack.append(int(t))
            else:
                b, a = stack.pop(), stack.pop()
                if t == "+": stack.append(a + b)
                elif t == "-": stack.append(a - b)
                elif t == "*": stack.append(a * b)
                else: stack.append(trunc(a / b))
        return stack[0]

* 참고 링크 : https://dev.to/seanpgallivan/solution-evaluate-reverse-polish-notation-192l

* 설명 참고 링크 : https://www.youtube.com/watch?v=P8xYysvjLd0 

 C++ 코드

C ++ code

int evalRPN(vector<string>& tokens)
{
    stack<int>st;
    for(auto it : tokens)
    {
        if(it=="+" || it=="-" || it=="*" || it=="/"  )
        {
            int f = st.top();st.pop();
            int s = st.top();st.pop();

            if (it == "+")
              { int r = s + f;  st.push(r); }
            else if (it == "*")
              { int r = s * f;  st.push(r); }
            else if (it == "-")
              { int r = s - f;   st.push(r); }
            else if (it == "/")
              { int r = s / f;   st.push(r); }
        }
        else{ st.push( stoi(it)); }
    }
    return st.top();
}

* 참고 링크 : https://leetcode.com/problems/evaluate-reverse-polish-notation/discuss/1234191/c-easy-optimized-solution-150-evaluate-reverse-polish-notation

 출처

https://leetcode.com/problems/evaluate-reverse-polish-notation/