# Basic Calculator LeetCode Solution | Easy Approach Share:

Basic Calculator 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.

Accepted279,472Submissions694,573

Time: O(n)
Space: O(n)

### C++

``````class Solution {
public:
int calculate(string s) {
int ans = 0;
int num = 0;
int sign = 1;
stack<int> stack{{sign}};  // stack.top(): current env's sign

for (const char c : s)
if (isdigit(c))
num = num * 10 + (c - '0');
else if (c == '(')
stack.push(sign);
else if (c == ')')
stack.pop();
else if (c == '+' || c == '-') {
ans += sign * num;
sign = (c == '+' ? 1 : -1) * stack.top();
num = 0;
}

return ans + sign * num;
}
};
``````

### Java

``````
class Solution {
public int calculate(String s) {
int ans = 0;
int num = 0;
int sign = 1;
Deque<Integer> stack = new ArrayDeque<>(); // stack.peek(): current env's sign
stack.push(sign);

for (final char c : s.toCharArray())
if (Character.isDigit(c))
num = num * 10 + (c - '0');
else if (c == '(')
stack.push(sign);
else if (c == ')')
stack.pop();
else if (c == '+' || c == '-') {
ans += sign * num;
sign = (c == '+' ? 1 : -1) * stack.peek();
num = 0;
}

return ans + sign * num;
}
}
``````

### Python

``````
class Solution:
def calculate(self, s: str) -> int:
ans = 0
num = 0
sign = 1
stack = [sign]  # stack[-1]: current env's sign

for c in s:
if c.isdigit():
num = num * 10 + (ord(c) - ord('0'))
elif c == '(':
stack.append(sign)
elif c == ')':
stack.pop()
elif c == '+' or c == '-':
ans += sign * num
sign = (1 if c == '+' else -1) * stack[-1]
num = 0

return ans + sign * num
``````

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