'''
中綴轉(zhuǎn)換為后綴
'''

a = "1+(2+6/1+2)"
#a = "2+9/3-5"
# 可能出現(xiàn)的符號 
symbol_1 = ['+','-','*','/']
symbol_2 = ['(']
symbol_3 = [')']
# 符號的優(yōu)先級
priority = {'#':-1,'(':1,'+':2,'-':2,'*':3,'/':3}
match_2 = {')':'('}
# 存儲符號的棧
zhan = []
zhan.append("#")
# 結(jié)果
result = []

'''
### 這里只是對表達(dá)式進(jìn)行轉(zhuǎn)換
for i in a:
 # 如果是數(shù)字直接添加到結(jié)果
 if i.isdigit():
 result.append(i)
 # 如果是 + - * / 運算，則先出棧更低優(yōu)先級的，然后入棧
 elif i in symbol_1:
 # 如果優(yōu)先級低，則出棧所有優(yōu)先級>=的符號
 while priority[i] = priority[zhan[-1]]:
 result.append(zhan.pop())
 # 壓入符號
 zhan.append(i)
 # 如果是左括號，直接壓入
 elif i in symbol_2:
 zhan.append(i)
 # 如果是右括號，則出棧，直到遇到了匹配的左括號，然后吧左括號也出棧
 elif i in symbol_3:
 while zhan[-1] != match_2[i]:
 result.append(zhan.pop())
 zhan.pop()
 
result.append(zhan.pop())
print(''.join(result))

'''
# 下面通過將中綴表達(dá)式轉(zhuǎn)換為后綴表達(dá)式，并進(jìn)行運算
def my_operation(symbol, a, b):
 a,b = int(a),int(b)
 if symbol == '+':
 return a + b
 elif symbol == '-':
 return a - b
 elif symbol == '*':
 return a * b
 elif symbol == '/':
 return a / b

def to_operation(result, zhan):
 two = result.pop()
 one = result.pop()
 symbol = zhan.pop()
 ret = my_operation(symbol, one, two)
 print(f"{one}{symbol}{two} = {ret}")
 result.append(ret)

### 在表達(dá)式轉(zhuǎn)換的時候就一邊進(jìn)行了運算
for i in a:
 # 如果是數(shù)字直接添加到結(jié)果
 if i.isdigit():
 result.append(i)
 # 如果是 + - * / 運算，則先出棧更低優(yōu)先級的，然后入棧
 elif i in symbol_1:
 # 如果優(yōu)先級低，則出棧所有優(yōu)先級>=的符號
 while priority[i] = priority[zhan[-1]]:
 to_operation(result, zhan)
 # 壓入符號
 zhan.append(i)
 # 如果是左括號，直接壓入
 elif i in symbol_2:
 zhan.append(i)
 # 如果是右括號，則出棧，直到遇到了匹配的左括號，然后吧左括號也出棧
 elif i in symbol_3:
 while zhan[-1] != match_2[i]:
 to_operation(result, zhan)
 zhan.pop()
 
to_operation(result, zhan)
print(result)