224. Basic Calculator

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本文介绍了一种使用栈数据结构实现的基本计算器算法，用于解析和计算包含加减运算符及括号的简单数学表达式。通过迭代输入字符串，算法能够处理非负整数和空格，确保了计算过程的正确性和效率。

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Description

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.

Problem URL

Solution

给一个String代表的只有+ - （）的表达式，保证合法，计算它的结果。

Using a stack to maintain the result before a () bracket. We use initial sign = 1. every time we meet a non-digit char, it is the time to calculate previous number value to result. It is calculating previour number and assigning number and sing value for next number. After the iteration, if we still have unadded number, remember add it.

Code

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