79 lines
2.4 KiB
Python
79 lines
2.4 KiB
Python
"""
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1. 经观察可发现操作数的相对位置是不变的，变化的只是操作符
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2. 中转前：
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（1）。 如果是一个操作数，则看一眼操作符栈中的栈顶是否是 */
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（1.1）. 如果是 */，则先拿到这个操作数，再从操作数栈弹出一个数
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再弹出操作符，拼接后放入操作数栈顶
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（1.2）。如果是 +-，则将此操作数入操作数栈
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（2）. 如果是一个 ( + - * /，直接入操作符栈顶
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（3）. 如果是 )，则执行出栈，直到遇到 (，出栈的步骤跟 （1.1）
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3. 中转后：
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（1）.
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"""
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from .stack import Stack
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def infix_to_prefix(infix: str):
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op_stack = Stack[str]()
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num_stack = Stack[str]()
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for item in infix:
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if item in ['(', '*', '+', '-', '/']:
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op_stack.push(item)
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elif '0' <= item <= 'z':
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if op_stack.peek() in ['*', '/']:
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temp = num_stack.pop()
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op = op_stack.pop()
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num_stack.push(op + temp + item)
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else:
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num_stack.push(item)
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elif item == ')':
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op = op_stack.pop()
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while op != '(':
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temp1 = num_stack.pop()
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temp2 = num_stack.pop()
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num_stack.push(op + temp2 + temp1)
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op = op_stack.pop()
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if not op_stack.is_empty():
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i1 = num_stack.pop()
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i2 = num_stack.pop()
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op = op_stack.pop()
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return op + i2 + i1
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else:
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return num_stack.pop()
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def infix_to_postfix(infix: str):
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op_stack = Stack[str]()
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num_stack = Stack[str]()
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for item in infix:
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if item in ['(', '*', '+', '-', '/']:
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op_stack.push(item)
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elif '0' <= item <= 'z':
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if op_stack.peek() in ['*', '/']:
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temp = num_stack.pop()
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op = op_stack.pop()
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num_stack.push(temp + item + op)
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else:
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num_stack.push(item)
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elif item == ')':
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op = op_stack.pop()
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while op != '(':
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temp1 = num_stack.pop()
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temp2 = num_stack.pop()
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num_stack.push(temp2 + temp1 + op)
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op = op_stack.pop()
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if not op_stack.is_empty():
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i1 = num_stack.pop()
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i2 = num_stack.pop()
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op = op_stack.pop()
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return i2 + i1 + op
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else:
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return num_stack.pop()
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res = infix_to_prefix('a+b*c')
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print(res)
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