How to use stack to implement simple calculator function in Java

1. Idea analysis

When we enter a simple infix expression similar to "7*2 100-5 3-4/2", our compiler can use the code we wrote The code scans this expression and calculates the result

There are two main elements in this expression, one is a number and the other is a symbol, so we need to create two stack structures to store the data

• Number stack numStack: stores numbers

• Symbol stack operStack: stores operators

##1. First we need to define an index to traverse our expression

2. If a number is scanned, it will be directly put into the stack

3. If an operator is scanned, Then there are the following situations:

3.1. If the current symbol stack is empty, push it directly into the stack

3.2. If the symbol stack has operators, it needs to be compared

If the priority of the current operator is less than or equal to the operator in the stack, you need to pop two numbers from the number stack, pop a symbol from the symbol stack, perform operations, and get The result is put into the number stack, and then the current operator is put into the symbol stack

If the priority of the current operator is greater than the operator on the stack, it is put directly into the symbol stack

4. When the expression scan is completed, the corresponding numbers and symbols are sequentially popped out from the number stack and symbol stack, and calculated

5. The last number retained in the number stack is the operation The result

2. Code implementation

package com.hsy.stack;

public class Calculator {

    public static void main(String[] args) {
        //根据前面老师思路，完成表达式的运算
        String expression = "7*2+100-5+3-4/2";//如何处理多位数的问题？
        //创建两个栈，数栈，一个符号栈
        ArrayStack2 numStack = new ArrayStack2(10);
        ArrayStack2 operStack = new ArrayStack2(10);
        //定义需要的相关变量
        int index = 0;//用于扫描
        int num1 = 0;
        int num2 = 0;
        int oper = 0;
        int res = 0;
        char ch = ' '; //将每次扫描得到char保存到ch
        String keepNum = ""; //用于拼接 多位数
        //开始while循环的扫描expression
        while(true) {
            //依次得到expression 的每一个字符
            ch = expression.substring(index, index+1).charAt(0);
            //判断ch是什么，然后做相应的处理
            if(operStack.isOper(ch)) {//如果是运算符
                //判断当前的符号栈是否为空
                if(!operStack.isEmpty()) {
                    //如果符号栈有操作符，就进行比较,如果当前的操作符的优先级小于或者等于栈中的操作符,就需要从数栈中pop出两个数,
                    //在从符号栈中pop出一个符号，进行运算，将得到结果，入数栈，然后将当前的操作符入符号栈
                    if(operStack.priority(ch) <= operStack.priority(operStack.peek())) {
                        num1 = numStack.pop();
                        num2 = numStack.pop();
                        oper = operStack.pop();
                        res = numStack.cal(num1, num2, oper);
                        //把运算的结果如数栈
                        numStack.push(res);
                        //然后将当前的操作符入符号栈
                        operStack.push(ch);
                    } else {
                        //如果当前的操作符的优先级大于栈中的操作符， 就直接入符号栈.
                        operStack.push(ch);
                    }
                }else {
                    //如果为空直接入符号栈..
                    operStack.push(ch); // 1 + 3
                }
            } else { //如果是数，则直接入数栈

                //numStack.push(ch - 48); //? "1+3" &#39;1&#39; => 1
                //分析思路
                //1. 当处理多位数时，不能发现是一个数就立即入栈，因为他可能是多位数
                //2. 在处理数，需要向expression的表达式的index 后再看一位,如果是数就进行扫描，如果是符号才入栈
                //3. 因此我们需要定义一个变量 字符串，用于拼接

                //处理多位数
                keepNum += ch;

                //如果ch已经是expression的最后一位，就直接入栈
                if (index == expression.length() - 1) {
                    numStack.push(Integer.parseInt(keepNum));
                }else{

                    //判断下一个字符是不是数字，如果是数字，就继续扫描，如果是运算符，则入栈
                    //注意是看后一位，不是index++
                    if (operStack.isOper(expression.substring(index+1,index+2).charAt(0))) {
                        //如果后一位是运算符，则入栈 keepNum = "1" 或者 "123"
                        numStack.push(Integer.parseInt(keepNum));
                        //重要的!!!!!!, keepNum清空
                        keepNum = "";

                    }
                }
            }
            //让index + 1, 并判断是否扫描到expression最后.
            index++;
            if (index >= expression.length()) {
                break;
            }
        }

        //当表达式扫描完毕，就顺序的从 数栈和符号栈中pop出相应的数和符号，并运行.
        while(true) {
            //如果符号栈为空，则计算到最后的结果, 数栈中只有一个数字【结果】
            if(operStack.isEmpty()) {
                break;
            }
            num1 = numStack.pop();
            num2 = numStack.pop();
            oper = operStack.pop();
            res = numStack.cal(num1, num2, oper);
            numStack.push(res);//入栈
        }
        //将数栈的最后数，pop出，就是结果
        int res2 = numStack.pop();
        System.out.printf("表达式 %s = %d", expression, res2);
    }

}

//先创建一个栈,直接使用前面创建好
//定义一个 ArrayStack2 表示栈, 需要扩展功能
class ArrayStack2 {
    private int maxSize; // 栈的大小
    private int[] stack; // 数组，数组模拟栈，数据就放在该数组
    private int top = -1;// top表示栈顶，初始化为-1

    //构造器
    public ArrayStack2(int maxSize) {
        this.maxSize = maxSize;
        stack = new int[this.maxSize];
    }

    //增加一个方法，可以返回当前栈顶的值, 但是不是真正的pop
    public int peek() {
        return stack[top];
    }

    //栈满
    public boolean isFull() {
        return top == maxSize - 1;
    }
    //栈空
    public boolean isEmpty() {
        return top == -1;
    }
    //入栈-push
    public void push(int value) {
        //先判断栈是否满
        if(isFull()) {
            System.out.println("栈满");
            return;
        }
        top++;
        stack[top] = value;
    }
    //出栈-pop, 将栈顶的数据返回
    public int pop() {
        //先判断栈是否空
        if(isEmpty()) {
            //抛出异常
            throw new RuntimeException("栈空，没有数据~");
        }
        int value = stack[top];
        top--;
        return value;
    }
    //显示栈的情况[遍历栈]， 遍历时，需要从栈顶开始显示数据
    public void list() {
        if(isEmpty()) {
            System.out.println("栈空，没有数据~~");
            return;
        }
        //需要从栈顶开始显示数据
        for(int i = top; i >= 0 ; i--) {
            System.out.printf("stack[%d]=%d\n", i, stack[i]);
        }
    }
    //返回运算符的优先级，优先级是程序员来确定, 优先级使用数字表示
    //数字越大，则优先级就越高.
    public int priority(int oper) {
        if(oper == &#39;*&#39; || oper == &#39;/&#39;){
            return 1;
        } else if (oper == &#39;+&#39; || oper == &#39;-&#39;) {
            return 0;
        } else {
            return -1; // 假定目前的表达式只有 +, - , * , /
        }
    }
    //判断是不是一个运算符
    public boolean isOper(char val) {
        return val == &#39;+&#39; || val == &#39;-&#39; || val == &#39;*&#39; || val == &#39;/&#39;;
    }
    //计算方法
    public int cal(int num1, int num2, int oper) {
        int res = 0; // res 用于存放计算的结果
        switch (oper) {
            case &#39;+&#39;:
                res = num1 + num2;
                break;
            case &#39;-&#39;:
                res = num2 - num1;// 注意顺序
                break;
            case &#39;*&#39;:
                res = num1 * num2;
                break;
            case &#39;/&#39;:
                res = num2 / num1;
                break;
            default:
                break;
        }
        return res;
    }

}

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