python 逆波兰式

技巧：为简化代码，引入一个不存在的运算符#，优先级最低。置于堆栈底部

class Stack(object):
    '''堆栈'''
    def __init__(self):
        self._stack = []
        
    def pop(self):
        return self._stack.pop()
    
    def push(self, x):
        self._stack.append(x)

一、表达式无括号

def solve(bds):
    '''不带括号，引入#运算符'''
    pro = dict(zip('^*/+-#', [3,2,2,1,1,0]))
    out = []
    s = Stack()
    s.push('#')
    for x in bds:
        if x in '^*/+-':
            t = s.pop()
            while pro[x] <= pro[t]:
                out.append(t)
                t = s.pop()

            s.push(t)
            s.push(x)
        else:
            out.append(x)
        
    while not s.is_null():
        out.append(s.pop())
        
    return out[:-1]

bds1 = 'a+b/c^d-e'          # abcd^/+e-
print(bds1, ''.join(solve(bds1)))

二、表达式有括号

def solve(bds):
    '''带括号，引入#运算符'''
    pro = dict(zip('^*/+-#', [3,2,2,1,1,0]))
    out = []
    s = Stack()
    s.push('#')
    for x in bds:
        if x == '(':            # ①左括号 -- 直接入栈
            s.push(x)
        elif x == ')':          # ②右括号 -- 输出栈顶，直至左括号(舍弃)
            t = s.pop()
            while t != '(':
                out.append(t)
                t = s.pop()
        elif x in '^*/+-':      # ③运算符 -- 从栈顶开始，优先级不小于x的都依次弹出；然后x入栈
            while True:
                t = s.pop()
                if t == '(':    # 左括号入栈前优先级最高，而入栈后优先级最低！
                    s.push(t)
                    break
                if pro[x] <= pro[t]:
                    out.append(t)
                else:
                    s.push(t)
                    break
            s.push(x)
        else:                   # ④运算数 -- 直接输出
            out.append(x)
        
    while not s.is_null():
        out.append(s.pop())
        
    return out[:-1]
        
bds1 = 'a+b/c^d-e'          # abcd^/+e-
bds2 = '(a+b)*c-(d+e)/f'    # ab+c*de+f/-

print(bds1, ''.join(solve(bds1)))
print(bds2, ''.join(solve(bds2)))

三、根据后缀表达式求值

def solve5(bds):
    '''根据后缀表达式求值'''
    jishuan = {
        '^': lambda x,y: x**y,
        '*': lambda x,y: x*y,
        '/': lambda x,y: x/y,
        '+': lambda x,y: x+y,
        '-': lambda x,y: x-y
    }
    s = Stack()
    for x in bds:
        if x in '^*/+-':
            num2, num1 = s.pop(), s.pop()
            r = jishuan[x](float(num1), float(num2))
            s.push(r)
        else:
            s.push(x)

    return s.pop()

bds1 = '2+9/3^2-5'         # 2932^/+5-   -2
bds2 = '(1+2)*3-(4+5)/6'   # ab+c*de+f/- 7.5

print(bds1, '=', solve5(solve(bds1)))
print(bds2, '=', solve5(solve(bds2)))

#print(bds1, '=', eval(bds1))
print(bds2, '=', eval(bds2))