0772. Basic Calculator III

https://leetcode.com/problems/basic-calculator-iii

Description

Implement a basic calculator to evaluate a simple expression string.

The expression string contains only non-negative integers, '+', '-', '*', '/' operators, and open '(' and closing parentheses ')'. The integer division should truncate toward zero.

You may assume that the given expression is always valid. All intermediate results will be in the range of [-231, 231 - 1].

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 = "6-4/2"
**Output:** 4

Example 3:

**Input:** s = "2*(5+5*2)/3+(6/2+8)"
**Output:** 21

Example 4:

**Input:** s = "(2+6*3+5-(3*14/7+2)*5)+3"
**Output:** -12

Example 5:

**Input:** s = "0"
**Output:** 0

Constraints:

• 1 <= s <= 104

• s consists of digits, '+', '-', '*', '/', '(', and ')'.

• s is a valid expression.

ac: Stack

class Solution {
    public int calculate(String s) {
        // Edge cases
        if (s == null || s.length() == 0) return 0;
        
        Deque<Character> operators = new ArrayDeque<>();
        Deque<Integer> nums = new ArrayDeque<>();
        
        int currNum = 0;
        for (int i = 0; i < s.length(); i++) {
            char c = s.charAt(i);
            if (c == ' ') continue;
            if (Character.isDigit(c)) {
                currNum = currNum * 10 + (c - '0');
            } else if (c == '+' || c == '-' || c == '*' || c == '/') {
                while (!operators.isEmpty() && canDoPreviousOperation(operators.peek(), c)) {
                    currNum = doOperation(nums.pop(), operators.pop(), currNum);
                }
                
                operators.push(c);
                nums.push(currNum);
                currNum = 0;
            } else if (c == '(') {
                operators.push('(');
            } else if (c == ')') {
                while (operators.peek() != '(') {
                    currNum = doOperation(nums.pop(), operators.pop(), currNum);
                }
                operators.pop(); // evict '('
            }
        }
        
        while (!operators.isEmpty()) {
            currNum = doOperation(nums.pop(), operators.pop(), currNum);
        }
        
        return currNum;
    }
    
    private int doOperation(int num1, Character operator, int num2) {
        switch (operator) {
            case '+': return num1 + num2;
            case '-': return num1 - num2;
            case '*': return num1 * num2;
            case '/': return num1 / num2;
            default:
                throw new IllegalArgumentException("Invalid operator.");
        }
    }
    
    private boolean canDoPreviousOperation(Character op1, Character op2) {
        if (op1 == '(') return false;
        if (op2 == '+' || op2 == '-') {
            return true;
        } else {
            // op2 is * or /
            if (op1 == '*' || op1 == '/') return true; 
            
            return false;
        }
        
    }
}
/*
E.g. ...(6 - 4 / 2)...
num1, op1, currNum, {currOp}, {num3} ...
1. currOp is +/-, or op1 && currOp both are * or /: currNum = eval(num1, op1, currNum)
2. other cases: currNum = eval(currNum, currOp, num3). Because we don't know whether it is left parenthesis or num3 after currOp, let's push currNum, currOp to stack and do calculation later.
stack operators: [-]
stack nums: [6]
*/