LeetCode No.150 | StriveZs的博客

LeetCode No.150

LeetCode第150题—逆波兰表达式求值

自己代码的开源仓库:click here 欢迎Star和Fork :)

题目描述

根据 逆波兰表示法,求表达式的值。

有效的算符包括 +、-、*、/ 。每个运算对象可以是整数,也可以是另一个逆波兰表达式。

说明:

  • 整数除法只保留整数部分。
  • 给定逆波兰表达式总是有效的。换句话说,表达式总会得出有效数值且不存在除数为 0 的情况。
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示例 1

输入:tokens = ["2","1","+","3","*"]
输出:9
解释:该算式转化为常见的中缀算术表达式为:((2 + 1) * 3) = 9
示例 2

输入:tokens = ["4","13","5","/","+"]
输出:6
解释:该算式转化为常见的中缀算术表达式为:(4 + (13 / 5)) = 6
示例 3

输入:tokens = ["10","6","9","3","+","-11","*","/","*","17","+","5","+"]
输出:22
解释:
该算式转化为常见的中缀算术表达式为:
((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


提示:

1 <= tokens.length <= 104
tokens[i] 要么是一个算符("+""-""*""/"),要么是一个在范围 [-200, 200] 内的整数

代码

可以说只要想到用栈和理解逆波兰式的含义就十分简单了。唯一一点就是要注意只取整数部分,不考虑小数部分。

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import math
class Solution(object):
def evalRPN(self, tokens):
"""
逆波兰式是后缀表达式
后缀表达式的求法:只需要处理符号的前两个值就可以了
那么就可以考虑使用栈来模拟计算过程
需要注意的是,这里只取整数部分,不考虑小数 因此考虑使用math.modf()
:type tokens: List[str]
:rtype: int
"""
stack = [] # 数字栈
characters = ['-', '+', '*', '/'] # 符号
for i in range(len(tokens)):
if tokens[i] in characters:
b = stack[-1]
stack.pop()
a = stack[-1]
stack.pop()
if tokens[i] == '-':
c = a - b
elif tokens[i] == '+':
c = a + b
elif tokens[i] == '*':
c = a * b
elif tokens[i] == '/':
c = a / b
stack.append(math.modf(c)[1])
else:
stack.append(int(tokens[i]))
return int(stack[-1])

if __name__ == '__main__':
s = Solution()
print(s.evalRPN(["10","6","9","3","+","-11","*","/","*","17","+","5","+"]))
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