思路

代码（可以看到这里的数字只能是单位数字，那么如何改成可以是多位数呢？！往下看）

package stack;public class Calculator {public static void main(String[] args) {//完成表达式运算String expression = "7+2*6-2*2";//创建两个栈，一个数栈，一个符号栈ArrayStack1 numstack = new ArrayStack1(10);ArrayStack1 operstack = new ArrayStack1(10);//定义需要的相关变量int index = 0; //用于扫描int num1 = 0, num2 = 0;int oper = 0;int res = 0;char ch = ' ';//将每次扫描得到的char保存给ch//开始while循环的扫描expressionwhile(true){//一次得到expression 的每个字符ch = expression.substring(index,index+1).charAt(0);//判断ch是字符还是数字if (operstack.isOper(ch)){//判断符号栈是否为空if (!operstack.isEmpty()){//如果不为空，比较，如果当前操作符小于等于则进行下面操作if (operstack.priority(ch) <= operstack.priority(operstack.peekTop())){num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);//把运算的结果加入栈numstack.push(res);//然后将当前操作符入符号栈operstack.push(ch);}else {operstack.push(ch);}}else {//为空直接入栈operstack.push(ch);}}else {//如果是数直接入栈(这里的1是字符'1'，要转成数字1，ascll码减48)numstack.push(ch - 48);}//让index + 1，判断是否扫描到expression最后index ++;if (index >= expression.length()){break;}}//扫描表达式完毕后，就顺序的从 数栈和符号栈中pop出相应的数和符合，并运算while (true){//如果符号栈为空，则计算得到最后的结果，数栈中只有一个数字if (operstack.isEmpty()){break;}num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);numstack.push(res);}//打印结果(就是数栈中的最后一个数)System.out.printf("表达式 %s = %d",expression,numstack.pop());}
}class ArrayStack1{private int maxSize; //栈大小private int[] stack;//数组模拟栈，数据放入数组里private int top = -1; //top表示栈顶，初始化为-1//构造器public ArrayStack1(int maxSize) {this.maxSize = maxSize;stack = new int[this.maxSize];}//返回栈顶的值，不出栈public int peekTop(){return stack[top];}//栈满public boolean isFull(){return top == maxSize - 1;}//栈空public boolean isEmpty(){return top == -1;}// 入栈 pushpublic void push(int value){//判断栈是否满if (isFull()){System.out.println("栈满，不能入栈");return;}top++;stack[top] = value;}//出栈public int pop(){if (isEmpty()){//抛出异常throw new RuntimeException("栈空");}int value = stack[top];top--;return value;}//遍历栈(从栈顶开始)public void list(){if (isEmpty()){System.out.println("栈空");}for (int i = top; i>=0; i--){System.out.printf("stack[%d] = %d\n",i,stack[i]);}}//返回运算符优先级，优先级有程序员来确定，优先级使用数字表示//数字越大，优先级越高public int priority(int oper){if (oper == '*' || oper == '/'){return 1;}else if(oper == '+' || oper == '-'){return 0;}else {return -1;//假定目前的表达式只有 +-*/}}//判断是否是一个运算符public boolean isOper(char val){return val == '+' || val == '-' || val == '*' || val == '/';}//计算方法public int cal(int num1, int num2, int oper){int res = 0; //计算结果switch (oper){case '+':res = num1 + num2;break;case '-':res = num2 - num1; //注意顺序break;case '*':res = num1 * num2;break;case '/':res = num2 / num1;break;}return res;}
}

找到是哪个地方有问题？，发现是一个数字就直接入栈了，那么需要在入数栈时，需要向后看，如果是数就拼接并继续扫描，如果是符合才入栈。（那么就需要一个字符串变量用于拼接）

代码

package stack;public class Calculator {public static void main(String[] args) {//完成表达式运算String expression = "71+2*6-2*2";//创建两个栈，一个数栈，一个符号栈ArrayStack1 numstack = new ArrayStack1(10);ArrayStack1 operstack = new ArrayStack1(10);//定义需要的相关变量int index = 0; //用于扫描int num1 = 0, num2 = 0;int oper = 0;int res = 0;char ch = ' ';//将每次扫描得到的char保存给chString keepnum = ""; //用于拼接多位数//开始while循环的扫描expressionwhile(true){//一次得到expression 的每个字符ch = expression.substring(index,index+1).charAt(0);//判断ch是字符还是数字if (operstack.isOper(ch)){//判断符号栈是否为空if (!operstack.isEmpty()){//如果不为空，比较，如果当前操作符小于等于则进行下面操作if (operstack.priority(ch) <= operstack.priority(operstack.peekTop())){num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);//把运算的结果加入栈numstack.push(res);//然后将当前操作符入符号栈operstack.push(ch);}else {operstack.push(ch);}}else {//为空直接入栈operstack.push(ch);}}else {//如果是数直接入栈(这里的1是字符'1'，要转成数字1，ascll码减48)//如果有多位数，则需要改进代码，不能直接入栈//思路：// 1.扫描到数字，需要继续向后扫描，如果是数就拼接，继续扫描，直到扫描是字符keepnum += ch;//如果ch是最后一位了则直接入栈if (index == expression.length()-1){numstack.push(Integer.parseInt(keepnum));}else {//判断下一个字符是不是数字，如果是继续扫描if (operstack.isOper(expression.substring(index + 1, index + 2).charAt(0))) {numstack.push(Integer.parseInt(keepnum));//这里一定要清空keepnumkeepnum = "";}}}//让index + 1，判断是否扫描到expression最后index ++;if (index >= expression.length()){break;}}//扫描表达式完毕后，就顺序的从 数栈和符号栈中pop出相应的数和符合，并运算while (true){//如果符号栈为空，则计算得到最后的结果，数栈中只有一个数字if (operstack.isEmpty()){break;}num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);numstack.push(res);}//打印结果(就是数栈中的最后一个数)System.out.printf("表达式 %s = %d",expression,numstack.pop());}
}class ArrayStack1{private int maxSize; //栈大小private int[] stack;//数组模拟栈，数据放入数组里private int top = -1; //top表示栈顶，初始化为-1//构造器public ArrayStack1(int maxSize) {this.maxSize = maxSize;stack = new int[this.maxSize];}//返回栈顶的值，不出栈public int peekTop(){return stack[top];}//栈满public boolean isFull(){return top == maxSize - 1;}//栈空public boolean isEmpty(){return top == -1;}// 入栈 pushpublic void push(int value){//判断栈是否满if (isFull()){System.out.println("栈满，不能入栈");return;}top++;stack[top] = value;}//出栈public int pop(){if (isEmpty()){//抛出异常throw new RuntimeException("栈空");}int value = stack[top];top--;return value;}//遍历栈(从栈顶开始)public void list(){if (isEmpty()){System.out.println("栈空");}for (int i = top; i>=0; i--){System.out.printf("stack[%d] = %d\n",i,stack[i]);}}//返回运算符优先级，优先级有程序员来确定，优先级使用数字表示//数字越大，优先级越高public int priority(int oper){if (oper == '*' || oper == '/'){return 1;}else if(oper == '+' || oper == '-'){return 0;}else {return -1;//假定目前的表达式只有 +-*/}}//判断是否是一个运算符public boolean isOper(char val){return val == '+' || val == '-' || val == '*' || val == '/';}//计算方法public int cal(int num1, int num2, int oper){int res = 0; //计算结果switch (oper){case '+':res = num1 + num2;break;case '-':res = num2 - num1; //注意顺序break;case '*':res = num1 * num2;break;case '/':res = num2 / num1;break;}return res;}
}