[LeetCode]Evaluate Reverse Polish Notation

最新推荐文章于 2025-08-14 22:07:01 发布

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最新推荐文章于 2025-08-14 22:07:01 发布

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文章标签： leetcode python stack

本文链接：https://blog.youkuaiyun.com/myself9711/article/details/40427179

本文介绍了一种利用栈的数据结构解决逆波兰表达式求值问题的方法，通过遍历表达式中的每个元素并进行相应的运算操作，实现了有效的括号处理。

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

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

Some examples:

  ["2", "1", "+", "3", "*"] -> ((2 + 1) * 3) -> 9
  ["4", "13", "5", "/", "+"] -> (4 + (13 / 5)) -> 6

Solution: Use Stack (mostly for Parentheses processing )

Running Time: O(n)

class Solution:
    # @param tokens, a list of string
    # @return an integer
    def isInt(self, s):
        try:
            int(s)
            return True
        except ValueError:
            return False
            
    
    def evalRPN(self, tokens):
        tempStack = []
        for element in tokens:
            if self.isInt(element):
                tempStack.append(int(element))
            else:
                b = tempStack.pop()
                a = tempStack.pop()
                if element == '+':
                    c = a + b
                elif element == '-':
                    c = a - b
                elif element == '*':
                    c = a * b
                elif element == '/':
                    try:
                        c = int(float(a) / b)#in python -6/132 = -1, but the answer here require 0 as result
                    except ValueError:
                        print "Error: divisor cannot be 0."
                tempStack.append(c)
        return int(tempStack.pop())

对于括号处理，通常使用Stack作为容器。