# Basic Calculator Problem

## Description

LeetCode Problem 224.

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:

``````1
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Input: s = "1 + 1"
Output: 2
``````

Example 2:

``````1
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Input: s = " 2-1 + 2 "
Output: 3
``````

Example 3:

``````1
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Input: s = "(1+(4+5+2)-3)+(6+8)"
Output: 23
``````

Constraints:

• 1 <= s.length <= 3 * 10^5
• 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.

## Sample C++ Code

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class Solution {
public:
int calculate(string s) {
// the given expression is always valid, only + and -
// every + and - can be flipped base on it's depth in ().
stack<int> signs;
int sign = 1;
int num = 0;
int ans = 0;

// always transform s into ( s )
signs.push(1);

for (auto c : s) {
if (c >= '0' && c <= '9') {
num = 10 * num + c - '0';
} else if (c == '+' || c == '-') {
ans = ans + signs.top() * sign * num;
num = 0;
sign = (c == '+' ? 1 : -1);
} else if (c == '(') {
signs.push(sign * signs.top());
sign = 1;
} else if (c == ')') {
ans = ans + signs.top() * sign * num;
num = 0;
signs.pop();
sign = 1;
}
}

if (num) {
ans = ans + signs.top() * sign * num;
}

return ans;
}
};
``````