772 Basic Calculator III

772. Basic Calculator III

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 theevalbuilt-in library function.

2. Implementation

(1) Stack

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, 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)) {
                // Handle negative number by pushing 0 to the stack
                // so that we can get negative number in the calculation. For example 0 - 2 = -2
                if (c == '-' && (i == 0 || s.charAt(i - 1) == '(')) {
                    nums.push(0);
                }
                
                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;
    }
    
    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)