# Leetcode: Basic Calculator II

Basic Calculator II Similar Problems:

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, +, -, *, / operators and empty spaces . The integer division should truncate toward zero.

You may assume that the given expression is always valid.

```Some examples:
"3+2*2" = 7
" 3/2 " = 1
" 3+5 / 2 " = 5
Note: Do not use the eval built-in library function.
```

Github: code.dennyzhang.com

Credits To: leetcode.com

Leave me comments, if you have better ways to solve.

```## Blog link: https://code.dennyzhang.com/basic-calculator-ii
## Basic Ideas: +- yield to */
##
##           When we found one operator as +/, look for the next operator
##           When we found one operator as */, we solve it immediately
## Complexity: Time O(n), Space O(n)
class Solution(object):
def calculate(self, s):
"""
:type s: str
:rtype: int
"""
s = s.replace(' ', '')
i = 0
length = len(s)
# solve */
stack = []
while i<length:
if s[i].isdigit():
# get the num
num_str = ''
while i<length and s[i].isdigit():
num_str = "%s%s" % (num_str, s[i])
i += 1
stack.append(num_str)
elif s[i] in '*/':
num1 = int(stack.pop())
num_str = ''
op = s[i]
# find the next number
i += 1
while i<length and s[i].isdigit():
num_str = "%s%s" % (num_str, s[i])
i += 1
num2 = int(num_str)
if op == '*':
num1 = num1*num2
else:
num1 = num1/num2
stack.append(str(num1))
else:
# +-
stack.append(s[i])
i += 1

# solve +-
res, i = 0, 0
while i<len(stack):
element = stack[i]
if element in '+-':
num2 = stack[i+1]
i = i+2
if element == '+':
res += int(num2)
else:
res -= int(num2)
else:
res += int(element)
i += 1
return res

# s = Solution()
# print s.calculate(" 3+5 / 2 ") # 5
```

Share It, If You Like It.