综合计算器怎么搞？
思路
Demo1
package datastructres.stack;
/**
* @author ：Yan Guang
* @date ：Created in 2021/1/11 9:41
* @description：
*/
public class Calculator {
public static void main(String[] args) {
String expression = "3+2*6-2";
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 = ' ';
while (true){
ch = expression.substring(index,index+1).charAt(0);
//首先判断ch是什么，然后进行操作
if (operStack.isOper(ch)){
if (!operStack.isEmpty()){
//如果该元素优先级小于等于栈顶的优先级
if (operStack.priority(ch)<=operStack.priority(operStack.peek())){
num1 = numStack.pop();
num2 = numStack.pop();
oper = operStack.pop();
res = operStack.cal(num1, num2, oper);
numStack.push(res);
operStack.push(ch);
}else {
//入栈
operStack.push(ch);
}
}else {
//如果为空，直接入栈
operStack.push(ch);
}
}else{
//因为numStack里面存放的是int类型的，这里存放值的时候需要对照ASCII码表进行对照
//但是int类型可以与char类型直接进行比较
numStack.push(ch-48);
}
index++;
if (index>=expression.length()){
break;
}
}
//现在扫描已经结束，需要继续处理完两个栈中的元素
while (true){
if (operStack.isEmpty()){
break;
}
num1 = numStack.pop();
num2 = numStack.pop();
oper = operStack.pop();
res = operStack.cal(num1, num2, oper);
numStack.push(res);
}
//最后存放在numStack里面的元素就是结果了
System.out.printf("表达式%s = %d",expression,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 boolean isFull() {
return top == maxSize - 1;
}
public boolean isEmpty() {
return top == -1;
}
public void push(int value) {
if (isFull()) {
System.out.println("栈满，无法添加~");
return;
}
top++;
stack[top] = value;
}
public int peek(){
return stack[top];
}
public int pop() {
if (isEmpty()) {
//throw相当于已经停止了代码，所以不用return
//当存在返回值的时候(不为void)的时候，我们尽量可以抛出一个异常来代替return
throw new RuntimeException("栈空，无法取出数据~");
}
int value = stack[top];
top--;
return value;
}
public void showStack() {
if (isEmpty()) {
System.out.println("栈已空~");
return;
}
//循环的时候是从栈顶往下循环，也就是从数组的最后开始循环的~
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;
default:
break;
}
return res;
}
}
这里存在的问题就是只能进行一位数运算，无法计算大于10的数，而且只有加减乘除四个运算符~
下面是优化之后的代码，具体也就改了一个地方，有很多注释的地方，大家看看吧
Demo2
package datastructres.stack;
/**
* @author ：Yan Guang
* @date ：Created in 2021/1/11 9:41
* @description：
*/
public class Calculator {
public static void main(String[] args) {
String expression = "70+2*6-4";
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="";
while (true){
ch = expression.substring(index,index+1).charAt(0);
//首先判断ch是什么，然后进行操作
if (operStack.isOper(ch)){
if (!operStack.isEmpty()){
//如果该元素优先级小于等于栈顶的优先级
if (operStack.priority(ch)<=operStack.priority(operStack.peek())){
num1 = numStack.pop();
num2 = numStack.pop();
oper = operStack.pop();
res = operStack.cal(num1, num2, oper);
numStack.push(res);
operStack.push(ch);
}else {
//入栈
operStack.push(ch);
}
}else {
//如果为空，直接入栈
operStack.push(ch);
}
}else{
//因为numStack里面存放的是int类型的，这里存放值的时候需要对照ASCII码表进行对照
//但是int类型可以与char类型直接进行比较
//1.
//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-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
// numStack.push(ch-48);
keepNum+=ch;
//如果当此时的数为最后一个数字数，就直接入栈
if (index==expression.length()-1){
numStack.push(Integer.parseInt(keepNum));
}else {
//这里会存在一个越界问题，如果index到了最后一个，就没有index+1了
if (operStack.isOper(expression.substring(index + 1, index + 2).charAt(0))) {//判断是否是运算符
numStack.push(Integer.parseInt(keepNum));
keepNum = "";
}
}
}
index++;
if (index>=expression.length()){
break;
}
}
//现在扫描已经结束，需要继续处理完两个栈中的元素
while (true){
if (operStack.isEmpty()){
break;
}
num1 = numStack.pop();
num2 = numStack.pop();
oper = operStack.pop();
res = operStack.cal(num1, num2, oper);
numStack.push(res);
}
//最后存放在numStack里面的元素就是结果了
System.out.printf("表达式%s = %d",expression,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 boolean isFull() {
return top == maxSize - 1;
}
public boolean isEmpty() {
return top == -1;
}
public void push(int value) {
if (isFull()) {
System.out.println("栈满，无法添加~");
return;
}
top++;
stack[top] = value;
}
public int peek(){
return stack[top];
}
public int pop() {
if (isEmpty()) {
//throw相当于已经停止了代码，所以不用return
//当存在返回值的时候(不为void)的时候，我们尽量可以抛出一个异常来代替return
throw new RuntimeException("栈空，无法取出数据~");
}
int value = stack[top];
top--;
return value;
}
public void showStack() {
if (isEmpty()) {
System.out.println("栈已空~");
return;
}
//循环的时候是从栈顶往下循环，也就是从数组的最后开始循环的~
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;
default:
break;
}
return res;
}
}
欢迎大家在评论区讨论哟~