栈实现计算器

public class Calculator {

    public static void main(String[] args) {
        String expressoin = "7*2*2-5+1-5+3-4";
        ArrayStack2 numStack = new ArrayStack2(10);
        ArrayStack2 operStack = new ArrayStack2(10);
        int index = 0;
        int num1 = 0;
        int num2 = 0;
        int oper = 0;
        int res = 0;
        char ch = ' ';
        String keepNum = "";//拼接多位数
        while (true) {
            ch = expressoin.substring(index, index + 1).charAt(0);
            if (operStack.isOper(ch)) {
                if (!operStack.isEmpty()) {
                    if (operStack.priority(ch) <= operStack.priority(operStack.peek())) {
                        num1 = numStack.pop();
                        num2 = numStack.pop();
                        oper = operStack.pop();
                        res = numStack.cal(num1, num2, oper);
                        numStack.push(res);
                        operStack.push(ch);
                    } else {
                        operStack.push(ch);
                    }
                } else {
                    operStack.push(ch);
                }
            } else {
//                numStack.push(ch - 48);
                //处理多位数
                keepNum += ch;
                //如果ch是expression最后一位则直接入栈
                if(index==expressoin.length()-1){
                    numStack.push(Integer.parseInt(keepNum));
                } else {
                    //判断下一个数字是不是数字 是扫描 不是入栈
                    if (operStack.isOper(expressoin.substring(index + 1, index + 2).charAt(0))) {
                        //是运算符 入栈
                        numStack.push(Integer.parseInt(keepNum));
                        keepNum = "";
                    }
                }
            }
            //让index+1 并判断是否扫描到expression最后
            index++;
            if (index >= expressoin.length()) {
                break;
            }
        }
        while (true) {
            if (operStack.isEmpty()) {
                break;
            }
            num1 = numStack.pop();
            num2 = numStack.pop();
            oper = operStack.pop();
            res = numStack.cal(num1, num2, oper);
            numStack.push(res);
        }
        System.out.println(expressoin + "=" + numStack.pop());
    }
}

class ArrayStack2 {
    private int maxSize;
    private int[] stack;
    private int top = -1;

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

    public int peek() {
        return stack[top];
    }

    public boolean isFull() {
        return top == maxSize - 1;
    }

    public boolean isEmpty() {
        return top == -1;
    }

    public void push(int value) {
        if (isFull()) {
            System.out.println("栈满");
            return;
        }
        top++;
        stack[top] = value;
    }

    public int pop() {
        if (isEmpty()) {
            throw new RuntimeException("栈空");
        }
        int value = stack[top];
        top--;
        return value;
    }

    public void list() {
        if (isEmpty()) {
            System.out.println("栈空");
            return;
        }
        for (int i = top; i >= 0; i--) {
            System.out.printf("stack[%d]=%d\n", i, stack[i]);
        }
    }

    public int priority(int oper) {
        if (oper == '*' || oper == '/') {
            return 1;
        } else if (oper == '+' || oper == '-') {
            return 0;
        } else {
            return -1;
        }
    }

    public boolean isOper(char val) {
        return val == '+' || val == '-' || val == '*' || val == '/';
    }

    public int cal(int num1, int num2, int oper) {
        int res = 0;
        switch (oper) {
            case '+':
                res = num1 + num2;
                break;
            case '-':
                res = num2 - num1;
                break;
            case '*':
                res = num1 * num2;
                break;
            case '/':
                res = num2 / num1;
                break;
        }
        return res;
    }
}
posted @ 2019-12-23 14:52  Axs  阅读(419)  评论(0编辑  收藏  举报