题目: https://leetcode.com/problems/basic-calculator/
难度: Medium
思路:
基本跟227一样,只是这里加了括号
瞄了一眼,基本上infix(中缀表达式)都是表达成postfix(后缀表达式)再来求值的。 比如 A + B * C 写成 A B C * +
Infix Expression | Prefix Expression | Postfix Expression |
---|---|---|
A + B | + A B | A B + |
A + B * C | + A * B C | A B C * + |
infix 中缀转postfix 后缀还有专门的算法:https://en.wikipedia.org/wiki/Shunting-yard_algorithm
-
Create an empty stack called opstack for keeping operators. Create an empty list for output.
-
Convert the input infix string to a list by using the string method split.
-
Scan the token list from left to right.
-
- If the token is an operand, append it to the end of the output list.
- If the token is a left parenthesis, push it on the opstack.
- If the token is a right parenthesis, pop the opstack until the corresponding left parenthesis is removed. Append each operator to the end of the output list.
- If the token is an operator, *, /, +, or -, push it on the opstack. However, first remove any operators already on the opstack that have higher or equal precedence and append them to the output list.
-
When the input expression has been completely processed, check the opstack. Any operators still on the stack can be removed and appended to the end of the output list.
可以看到中缀转后缀一个重要的点是: 当我们把operator +-*/ 放到opstack上时候,我们需要考虑/看是否有之前的operator有更高或者相等的precedence,这个时候我们需要优先(计算)把它放到output list.
参考
http://interactivepython.org/runestone/static/pythonds/BasicDS/InfixPrefixandPostfixExpressions.html
AC代码
class Solution(object):
def calculate(self, s):
"""
:type s: str
:rtype: int
"""
def precedence(op):
if op == '*' or op == '/':
return 2
else:
return 1
def cal(op, op1, op2):
if op == '*':
return op1 * op2
elif op == '/':
return op1 / float(op2)
elif op == '+':
return op1 + op2
else:
return op1 - op2
opstack = []
operands = []
# remove empty space and put operands and
idx = 0
for i in range(idx, len(s)):
if s[i] in '+-*/':
operands.append(s[idx:i])
while len(opstack) > 0 and precedence(s[i]) <= precedence(opstack[-1]) and len(operands) >= 2:
op = opstack.pop()
op2 = int(operands.pop())
op1 = int(operands.pop())
res = cal(op, op1, op2)
operands.append(res)
opstack.append(s[i])
idx = i + 1
operands.append(s[idx:])
while opstack:
op = opstack.pop()
op2 = int(operands.pop())
op1 = int(operands.pop())
res = cal(op, op1, op2)
operands.append(res)
return int(operands[0])