227. Basic Calculator II
Read the full problem statement on LeetCode.
Difficulty: medium Acceptance: 46% Topics: Math, String, Stack
View full problem on LeetCode Reading material
Reference solution (spoiler · python)
# Time: O(n)
# Space: O(n)
import operator
class Solution(object):
def calculate(self, s):
"""
:type s: str
:rtype: int
"""
def compute(operands, operators):
right, left = operands.pop(), operands.pop()
operands.append(ops[operators.pop()](left, right))
ops = {'+':operator.add, '-':operator.sub, '*':operator.mul, '/':operator.div}
precedence = {'+':0, '-':0, '*':1, '/':1}
operands, operators, operand = [], [], 0
for i in xrange(len(s)):
if s[i].isdigit():
operand = operand*10 + int(s[i])
if i == len(s)-1 or not s[i+1].isdigit():
operands.append(operand)
operand = 0
elif s[i] == '(':
operators.append(s[i])
elif s[i] == ')':
while operators[-1] != '(':
compute(operands, operators)
operators.pop()
elif s[i] in precedence:
while operators and operators[-1] in precedence and \
precedence[operators[-1]] >= precedence[s[i]]:
compute(operands, operators)
operators.append(s[i])
while operators:
compute(operands, operators)
return operands[-1]
# Time: O(n)
# Space: O(n)
class Solution2(object):
# @param {string} s
# @return {integer}
def calculate(self, s):
operands, operators = [], []
operand = ""
for i in reversed(xrange(len(s))):
if s[i].isdigit():
operand += s[i]
if i == 0 or not s[i-1].isdigit():
operands.append(int(operand[::-1]))
operand = ""
elif s[i] == ')' or s[i] == '*' or s[i] == '/':
operators.append(s[i])
elif s[i] == '+' or s[i] == '-':
while operators and \
(operators[-1] == '*' or operators[-1] == '/'):
self.compute(operands, operators)
operators.append(s[i])
elif s[i] == '(':
while operators[-1] != ')':
self.compute(operands, operators)
operators.pop()
while operators:
self.compute(operands, operators)
return operands[-1]
def compute(self, operands, operators):
left, right = operands.pop(), operands.pop()
op = operators.pop()
if op == '+':
operands.append(left + right)
elif op == '-':
operands.append(left - right)
elif op == '*':
operands.append(left * right)
elif op == '/':
operands.append(left / right)
Solution from kamyu104/LeetCode-Solutions · MIT
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