Given a string s representing a valid expression, implement a basic calculator to evaluate it, and return the result of the evaluation.
Note: You are not allowed to use any built-in function which evaluates strings as mathematical expressions, such as eval().
Example 1:
Input: s = "1 + 1"
Output: 2
Example 2:
Input: s = " 2-1 + 2 "
Output: 3
Example 3:
Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23
Constraints:
-
1 <= s.length <= 3 * 105 -
sconsists of digits,'+','-','(',')', and' '. -
srepresents a valid expression. -
'+'is not used as a unary operation (i.e.,"+1"and"+(2 + 3)"is invalid). -
'-'could be used as a unary operation (i.e.,"-1"and"-(2 + 3)"is valid). - There will be no two consecutive operators in the input.
- Every number and running calculation will fit in a signed 32-bit integer.
SOLUTION:
class Solution: def eval(self, a, op, b): if op == '+': return a + b elif op == '-': return a - b elif op == '/': return int(a / b) elif op == '*': return a * b def calculate(self, s: str) -> int: s += " " precedence = { '*': 1, '/': 1, '+': 0, '-': 0 } brackets = { '(': 1, ')': -1 } numstack = [] opstack = [] chunk = "" latestOpenBracket = True for c in s: if c == " " or c in precedence or c in brackets: if len(chunk) > 0: numstack.append(int(chunk)) chunk = "" if c in brackets: if brackets[c] == 1: opstack.append('(') latestOpenBracket = True elif brackets[c] == -1: latestOpenBracket = False while opstack[-1] != '(': currop = opstack.pop() b = numstack.pop() a = numstack.pop() res = self.eval(a, currop, b) numstack.append(res) opstack.pop() elif c in precedence: if latestOpenBracket: numstack.append(0) while len(opstack) > 0 and precedence.get(opstack[-1], -1) >= precedence[c]: currop = opstack.pop() b = numstack.pop() a = numstack.pop() res = self.eval(a, currop, b) numstack.append(res) opstack.append(c) latestOpenBracket = False else: chunk += c latestOpenBracket = False while len(opstack) > 0: currop = opstack.pop() b = numstack.pop() a = numstack.pop() res = self.eval(a, currop, b) numstack.append(res) return numstack[0] 
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