[LC] 224. Basic Calculator

最新推荐文章于 2022-02-10 04:39:45 发布

原创最新推荐文章于 2022-02-10 04:39:45 发布 · 328 阅读

0 ·

CC 4.0 BY-SA版权

LC解题报告专栏收录该内容

23 篇文章

订阅专栏

本文介绍了一种不使用内置eval函数实现基本计算器的方法，通过解析简单的数学表达式字符串并进行计算。文章详细讨论了如何利用栈来处理括号，并给出了具体的Java代码实现。

一、问题描述

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.

二、我的思路

想到在出现括号时用栈，但不知道java的栈怎么写……另外具体的边界条件也没想清楚。借鉴了discuss里的思路以后，写了代码

class Solution {
    public int calculate(String s) {
        char[] c = s.toCharArray();
        Stack<Integer> stack = new Stack<Integer>();
        
        int result = 0;
        int sign = 1;
        
        for(int i = 0; i < c.length; i ++){
            switch(c[i]){
                case '(':
                    stack.push(sign);
                    stack.push(result);
                    result = 0;
                    sign = 1;
                    break;
                case ')':
                    result = stack.pop() + stack.pop()*result;
                    break;
                case '+':
                    sign = 1;
                    break;
                case '-':
                    sign = -1;
                    break;
                case ' ':
                    break;  
                default:
                    if(Character.isDigit(c[i])){
                        int sum = c[i] - '0';
                        while(i + 1 <= c.length -1 && Character.isDigit(c[i+1])){	// Mind the condition!!!
                            sum = sum*10 + c[i + 1] - '0';
                            i ++;
                        }
                        result += sign * sum;
                    }
                    break;
            }
        }
        return result; 
    }
}

注意while的条件，以及sign和result什么时候复原。

三、淫奇技巧

思路都差不多，就是对五种字符分别处理，不同的是number的处理更优美了。

public int calculate(String s) {
    Stack<Integer> stack = new Stack<Integer>();
    int result = 0;
    int number = 0;
    int sign = 1;
    for(int i = 0; i < s.length(); i++){
        char c = s.charAt(i);
        if(Character.isDigit(c)){
            number = 10 * number + (int)(c - '0');
        }else if(c == '+'){
            result += sign * number;
            number = 0;
            sign = 1;
        }else if(c == '-'){
            result += sign * number;
            number = 0;
            sign = -1;
        }else if(c == '('){
            //we push the result first, then sign;
            stack.push(result);
            stack.push(sign);
            //reset the sign and result for the value in the parenthesis
            sign = 1;   
            result = 0;
        }else if(c == ')'){
            result += sign * number;  
            number = 0;
            result *= stack.pop();    //stack.pop() is the sign before the parenthesis
            result += stack.pop();   //stack.pop() now is the result calculated before the parenthesis
            
        }
    }
    if(number != 0) result += sign * number;
    return result;
}

四、不足

很多java的工具根本不知道……

Java Character：http://www.runoob.com/java/java-character.html

五、碎碎念

后天上午就要面试了，加油吖～