Basic Calculator

最新推荐文章于 2020-10-25 00:22:32 发布

原创最新推荐文章于 2020-10-25 00:22:32 发布 · 215 阅读

0 ·

CC 4.0 BY-SA版权

stack 同时被 2 个专栏收录

5 篇文章

订阅专栏

Math

2 篇文章

订阅专栏

本文介绍了一个简易计算器的设计与实现，该计算器能解析并计算包含括号、加减运算及空格的简单数学表达式。文中提供了具体的Java代码示例，展示了如何逐字符遍历输入字符串，利用栈来处理括号内的表达式。

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 .

You may assume that the given expression is always valid.

Some examples:

"1 + 1" = 2
" 2-1 + 2 " = 3
"(1+(4+5+2)-3)+(6+8)" = 23

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

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