【LeetCode with Python】 Evaluate Reverse Polish Notation

最新推荐文章于 2022-03-30 11:14:00 发布

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最新推荐文章于 2022-03-30 11:14:00 发布

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本文链接：https://blog.youkuaiyun.com/nerv3x3/article/details/3465981

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本文介绍了一种使用栈来计算逆波兰表达式的算法实现。通过分析输入字符串，该算法能够处理加、减、乘、除四种运算符，并正确计算出最终结果。文中提供了一个Python示例代码。

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博客域名： http://www.xnerv.wang
原题页面： https://oj.leetcode.com/problems/evaluate-reverse-polish-notation/
题目类型：栈
难度评价：★★
本文地址： http://blog.youkuaiyun.com/nerv3x3/article/details/3465981

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

逆波兰式的计算。相信大多数计算机和软件专业的朋友大学时都做过的算法题，用栈模拟计算过程。

class Solution:

    # @param tokens, a list of string
    # @return an integer
    def evalRPN(self, tokens):

        len_tokens = len(tokens)
        if 0 == len_tokens:
            return 0

        # python has no stack, so we have to simulate it
        nums_stack = []

        for i in range(0, len_tokens):
            token = tokens[i]
            # consider negative number, so shouldn't decide by the first char, but last char
            last_ch = token[len(token) - 1]
            if last_ch >= '0' and last_ch <= '9':
                nums_stack.append(int(token))
            else:
                right = nums_stack.pop()
                left = nums_stack.pop()
                if '+' == token:
                    result = left + right
                elif '-' == token:
                    result = left - right
                elif '*' == token:
                    result = left * right
                else:
                    sign = 1 if (left >= 0 and right >= 0) or (left <= 0 and right <= 0) else -1
                    result = abs(left) // abs(right) * sign
                nums_stack.append(result)

        return nums_stack.pop()