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224.basic-calculator.py
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119 lines (96 loc) · 2.75 KB
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# Tag: Math, String, Stack, Recursion
# Time: O(N)
# Space: O(N)
# Ref: -
# Note: -
# 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
# s consists of digits, '+', '-', '(', ')', and ' '.
# s represents 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.
#
#
class Solution:
def calculate(self, s: str) -> int:
stack = []
num = 0
sign = 1
result = 0
for char in s:
if char.isdigit():
num = num * 10 + int(char)
elif char in "+-":
result += sign * num
num = 0
sign = 1 if char == '+' else -1
elif char == "(":
stack.append(result)
stack.append(sign)
result = 0
sign = 1
elif char == ")":
result += sign * num
num = 0
result *= stack.pop()
result += stack.pop()
return result + sign * num
class Solution:
def calculate(self, s: str) -> int:
res, i = self.cal(s + '+', 0)
return res
def cal(self, s: str, i: int) -> int:
n = len(s)
pre, cur = 0, 0
op = None
while i < n:
while s[i] == ' ':
i += 1
num = 0
sign = 1
if s[i] == '-':
sign = -1
i += 1
while s[i].isdigit():
num = num * 10 + int(s[i])
i += 1
while s[i] == ' ':
i += 1
if s[i] == '(':
num, i = self.cal(s, i + 1)
num = num * sign
if op is None:
cur = num
elif op == '+':
pre = pre + cur
cur = num
elif op == '-':
pre = pre + cur
cur = -num
if s[i] == ')':
return pre + cur, i + 1
op = s[i]
i += 1
return pre + cur, i