# 772    Basic Calculator III

## 772. [Basic Calculator III](https://leetcode.com/problems/basic-calculator-iii/description/)

## 1. Question

Implement a basic calculator to evaluate a simple expression string.

The expression string may contain open`(`and closing parentheses`)`, the plus`+`or minus sign`-`,**non-negative**integers and empty spaces.

The expression string contains only non-negative integers,`+`,`-`,`*`,`/`operators , open`(`and closing parentheses`)`and empty spaces. 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`[-2147483648, 2147483647]`.

Some examples:

```
"1 + 1" = 2
" 6-4 / 2 " = 4
"2*(5+5*2)/3+(6/2+8)" = 21
"(2+6* 3+5- (3*14/7+2)*5)+3"=-12
```

**Note:Do not** use the`eval`built-in library function.

## 2. Implementation

**(1) Stack**

```java
class Solution {
    public int calculate(String s) {
        if (s == null || s.length() == 0) {
            return 0;
        }

        Stack<Integer> nums = new Stack();
        Stack<Character> operators = new Stack();

        int i = 0;
        int num = 0;
        int n = s.length();

        while (i < n) {
            char c = s.charAt(i);

            if (Character.isDigit(c)) {
                num = c - '0';

                while ((i + 1) < n && Character.isDigit(s.charAt(i + 1))) {
                    num = 10 * num + s.charAt(i + 1) - '0';
                    ++i;
                }
                nums.push(num);
            }
            else if (isOperator(c)) {
                while (!operators.isEmpty() && hasPrecedence(c, operators.peek())) {
                    nums.push(calculate(operators.pop(), nums.pop(), nums.pop()));
                }
                operators.push(c);
            }
            else if (c == '(') {
                operators.push(c);
            }
            else if (c == ')') {
                while (!operators.isEmpty() && operators.peek() != '(') {
                    nums.push(calculate(operators.pop(), nums.pop(), nums.pop()));
                }
                operators.pop();
            }
            ++i;
        }

        while (!operators.isEmpty()) {
            nums.push(calculate(operators.pop(), nums.pop(), nums.pop()));
        }
        return nums.isEmpty() ? 0 : nums.pop();
    }

    public boolean isOperator(char c) {
        return c == '+' || c == '-' || c == '*' || c == '/';
    }

    public int calculate(char operator, int num1, int num2) {
        int res = 0;

        switch(operator) {
            case '+':
                res = num1 + num2;
                break;
            case '-':
                res = num2 - num1;
                break;
            case '*':
                res = num1 * num2;
                break;
            case '/':
                res = num2 / num1;
                break;
        }
        return res;
    }

    // check if op2 has higher precendence than op1
    public boolean hasPrecedence(char op1, char op2) {
        if (op2 == ')' || op2 == '(') {
            return false;
        }

        if ((op1 == '*' || op1 == '/') && (op2 == '+' || op2 == '-')) {
            return false;
        }
        return true;
    }
}
```

## 3. Time & Space Complexity

**Stack:** 时间复杂度: O(n), 空间复杂度: O(n)