224. Basic Calculator

leetcode

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 {
public:
    int calculate(string s) {
        stack<int> stk;
        int res = 0;
        int sign = 1;
        int operand = 0;
        for(int i = 0; i < s.size(); i++) {
            char c = s[i];
            if(isdigit(c)) {
                operand = 10 * operand + (c - '0');
            } else if(c == '+') {
                res += sign * operand;
                sign = 1;
                // 已經將答案存到 res 了，operand reset 為 0
                operand = 0;
            } else if(c == '-') {
                res += sign * operand;
                sign = -1;
                // 已經將答案存到 res 了，operand reset 為 0
                operand = 0;
            } else if(c == '(') {
                // 將結果暫時存到 stack
                stk.push(res);

                // 也將 sign 暫時存到 stack
                stk.push(sign);

                // reset 結果為 0，重新計算
                res = 0;

                // reset sign 為 1，重新計算
                sign = 1;
            } else if(c == ')') {
                // 加上目前的計算
                res += sign * operand;

                // 乘上之前運算的結果的 sign
                res *= stk.top(); stk.pop();

                // 加上之前運算的結果
                res += stk.top(); stk.pop();

                // reset operand 為 0
                operand = 0;
            }
        }
        return res + (sign * operand);
    }
};
  • T: O(N)O(N)
  • S: O(N)O(N)