Basic Calculator LeetCode Solution | Easy Approach

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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

Basic Calculator Solutions

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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