Basic Calculator

Description

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.

Code

Time Complexity: $O(n)$, Space Complexity: $O(n)$

// order matters
sum += sign * num;
num = 0;

這段 code 必須放在 if else 判斷的前面，放在後面的話會產生錯誤的計算，以 "2-1 + 2" 為例，若放在後面，則 - 反而會作用在 2 身上，而非我們想要的 1 身上。

class Solution {
public:
    int calculate(string s) {
        stack<long long> nums, ops;
        long long num = 0;
        long long  sum = 0;
        long long  sign = 1;
        for(char c: s) {
            if(isdigit(c)) {
                num = num * 10 + c - '0';
            } else {
		        // order matters
                sum += sign * num;
                num = 0;
                
                if(c == '(') {
                    nums.push(sum);
                    ops.push(sign);
                    sum = 0;
                    sign = 1;
                }
                if(c == ')') {
                    sum = ops.top() * sum + nums.top();
                    nums.pop();
                    ops.pop();
                }
                if(c == '+') sign = 1;
                if(c == '-') sign = -1;
            }
        }
        sum += sign * num;
        return sum;
    }
};

Source

• Basic Calculator - LeetCode