实现栈的表达式的转化

偏执的太偏执、 2024-03-17 13:41 36阅读 0赞

## **1** **问题** ##

利用python实现栈的中缀表达式和后缀表达式的转变，可以解决简单的数学问题。

## **2** **方法** ##

使用def定义函数和类函数，循环函数等和一些基本的逻辑思维可以解决该问题。

代码清单 1

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 <td>from stack_set import Stack def infixToPostfix(infixpr): &nbsp; &nbsp; pre={} &nbsp; &nbsp; pre['*'],pre['/']=3,3 &nbsp; &nbsp; pre['+'],pre['-']=2,2 &nbsp; &nbsp; pre['(']=1 &nbsp; &nbsp; charlist=[chr(i) for i in range(65,91)] &nbsp; &nbsp; numlist=[i for i in range(10)] &nbsp; &nbsp; opstack=Stack() &nbsp; &nbsp; postfixlist=[] &nbsp; &nbsp; items=infixpr.split() &nbsp; &nbsp; for item in items: &nbsp; &nbsp; &nbsp; &nbsp; if item in charlist or item in numlist: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; postfixlist.append(item) &nbsp; &nbsp; &nbsp; &nbsp; elif item=="(": &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; opstack.push(item) &nbsp; &nbsp; &nbsp; &nbsp; elif item==')': &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tokentop=opstack.pop() &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; while tokentop!="(": &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; postfixlist.append(tokentop) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; tokentop=opstack.pop() &nbsp; &nbsp; &nbsp; &nbsp; else: &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; while(not opstack.isEmpty()) and (pre[item]&lt;=pre[opstack.peek()]): &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; postfixlist.append(opstack.pop()) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; opstack.push(item) &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; &nbsp; print("stack:{},postlist:{}".format(opstack.items,postfixlist)) &nbsp; &nbsp; while not opstack.isEmpty(): &nbsp; &nbsp; &nbsp; &nbsp; postfixlist.append(opstack.pop()) &nbsp; &nbsp; return "".join(postfixlist) if __name__ == '__main__': &nbsp; &nbsp; infix="A + ( B * C )" &nbsp; &nbsp; postfix=infixToPostfix(infix) &nbsp; &nbsp; print("其对应的后缀表达式：{}".format(postfix))</td> 
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## **3** **结语** ##

针对利用python来解决数学问题，提出循环定义等方法，通过实践，证明该方法是有效的。此方法结合并熟练的运用所学的简单知识和简单的逻辑关系，当然方法还有不足或考虑不周的地方，还有待继续研究。