Question

我已经实施了以下＆＃39;树大小的＆＃39;但它在某些条件下失败，下面的示例返回大小2时它应该返回大小4，任何人都可以帮助我。我已多次写这篇文章但无济于事，它一直都在失败。提前致谢

JC

def getRPNdepth(expression):
    treesize=0
    maxtreesize=treesize
    mintreesize=treesize
    tmpexp=expression
    tmpfmla = [1 if n[0] == 'x' else n for n in tmpexp]
    print(tmpfmla)
    try:
        stack = []
        for val in tmpfmla:
            if val in ['-', '+', '*', '/']:

                op1 = stack.pop()
                op2 = stack.pop()
                if val == '-': result = op2 - op1
                if val == '+': result = op2 + op1
                if val == '*': result = op2 * op1
                if val == '/':
                    if op1 == 0:
                        result = 1
                    else:
                        result = op2 / op1
                stack.append(result)
                treesize=treesize+1
            else:

                stack.append(float(val))
                treesize = treesize - 1

            if treesize>maxtreesize:
                maxtreesize=treesize
            if treesize<mintreesize:
                mintreesize=treesize
        return abs(mintreesize)
    except:
        print('error validate rpn>' + str(expression))
        return 0



xxxx = ['x6', 'x7', '+', 'x7', '+', 'x7', '+', 'x7', '+']
print(getRPNdepth(xxxx))

几个例子： [＆＃39; 1＆＃39;＆＃39; 1＆＃39; ＆＃39; +＆＃39;＆＃39; 1＆＃39;＆＃39; 1＆＃39; ＆＃39; +＆＃39; ＆＃39; +＆＃39;] [＆＃39; 1＆＃39;＆＃39; 1＆＃39;＆＃39; 1＆＃39; ＆＃39; +＆＃39; ＆＃39; +＆＃39;] 两者都给出结果为3，这是正确的，但是。 [＆＃39; 1＆＃39;＆＃39; 1＆＃39; ＆＃39; +＆＃39;＆＃39; 1＆＃39;＆＃39; 1＆＃39; ＆＃39; +＆＃39; ＆＃39; +＆＃39;] 它应该是4

时返回3

总而言之，我需要从字符串表示中知道RPN的深度。

Answer 1

计算树深度与评估表达式类似，但运算符计算结果深度而不是结果值：

def getRPNdepth(expression):
    stack = []
    for val in expression:
        if val in ['-', '+', '*', '/']:
            stack.append(max(stack.pop(),stack.pop())+1)
        else:
            stack.append(1)
    return stack.pop()

Answer 2

好吧，只是做了一些'作弊'并使用我的rpn到中缀转换器来实现相同的目标，如果有人需要它我在这里发布。

def getRPNdepth(expression):
    tmpexp = expression
    tmpfmla = [1 if n[0] == 'x' else n for n in tmpexp]
    stack = []
    for val in tmpfmla:
        if val!=' ':
            if val in ['-', '+', '*', '/']:
                op1 = stack.pop()
                op2 = stack.pop()
                stack.append('(' + str(op1) + str(val) + str(op2) + ')')
            else:
                stack.append(str(val))

    openparentesiscount=0
    maxopenparentesiscount = 0

    onlyparentesis=''
    for c in stack[0]:
        if c in ['(', ')']:
            onlyparentesis=onlyparentesis+c
            if c=='(':
                openparentesiscount=openparentesiscount+1
            else:
                openparentesiscount = openparentesiscount - 1
        if openparentesiscount>maxopenparentesiscount:
            maxopenparentesiscount=openparentesiscount

    return maxopenparentesiscount

全部谢谢！

获取RPN树大小

2 个答案: