224. Basic Calculator

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 .

 

Example 1:

Input: "1 + 1"
Output: 2

Example 2:

Input: " 2-1 + 2 "
Output: 3

Example 3:

Input: "(1+(4+5+2)-3)+(6+8)"
Output: 23

 

Note:

• You may assume that the given expression is always valid.
• Do not use the eval built-in library function.

 

Approach #1: Math. [Java]

class Solution {
    public int calculate(String s) {
        int ret = 0;
        Stack<Integer> stack = new Stack<>();
        
        int sign = 1;
        int number = 0;
        
        for (int i = 0; i < s.length(); ++i) {
            char c = s.charAt(i);
            if (Character.isDigit(c)) {
                number = number * 10 + c - '0';
            } else if (c == '-') {
                ret += number * sign;
                number = 0;
                sign = -1;
            } else if (c == '+') {
                ret += number * sign;
                number = 0;
                sign = 1;
            } else if (c == '(') {
                stack.push(ret);
                stack.push(sign);
                ret = 0;
                sign = 1;
            } else if (c == ')') {
                ret += number * sign;
                ret = ret * stack.pop() + stack.pop();
                number = 0;
                sign = 1;
            }
        }
        
        if (number != 0) ret += number * sign;
        
        return ret;
    }
}

　　

Analysis:

Simple iterative solution by identifying character one by one. One important thing is that the input is valid, which means the parenthese are always paired and in order.

Only 5 possible input we need to pay attention:

1. digit: it should be one digit from current number.

2. '+': number is over, we can add the the previous number and start a new number.

3. '-': same as above.

4. '(': push the previous result and the sign into the stack, set result to 0, just calculate the new result within the parenthesis.

5. ')': pop out the top two number from stack, first one is the sign before this pair of parenthesis, second is the temporary reslut before this pair of parenthesis. We add them together.

Finally if there is only one number, from the above solution, we haven't add the number to the result, so we do a check see if the number is zero.

 

Reference:

https://leetcode.com/problems/basic-calculator/discuss/62361/Iterative-Java-solution-with-stack