150-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
My solution：
class Solution(object): def evalRPN(self, tokens): “”” :type tokens: List[str] :rtype: int “”” stack = [] for token in tokens: if token == ‘+’ or token == ‘-‘ or token == ‘’ or token == ‘/’: a = int(stack.pop()) b = int(stack.pop()) res = 0 if token == ‘+’: res = a + b stack.append(res) elif token == ‘’: res = a * b stack.append(res) elif token == ‘/’: # truncates toward zero if b % a == 0: res = b / a else: res = b // a if a * b >= 0 else b // a + 1 stack.append(res) elif token == ‘-‘: res = b - a stack.append(res) else: stack.append(token)
return int(stack[0])
Time Complexity: O(n) Space Complexity: O(n)