思路

代碼（可以看到這里的數字只能是單位數字，那么如何改成可以是多位數呢？！往下看）

package stack;public class Calculator {public static void main(String[] args) {//完成表達式運算String expression = "7+2*6-2*2";//創建兩個棧，一個數棧，一個符號棧ArrayStack1 numstack = new ArrayStack1(10);ArrayStack1 operstack = new ArrayStack1(10);//定義需要的相關變量int index = 0; //用于掃描int num1 = 0, num2 = 0;int oper = 0;int res = 0;char ch = ' ';//將每次掃描得到的char保存給ch//開始while循環的掃描expressionwhile(true){//一次得到expression 的每個字符ch = expression.substring(index,index+1).charAt(0);//判斷ch是字符還是數字if (operstack.isOper(ch)){//判斷符號棧是否為空if (!operstack.isEmpty()){//如果不為空，比較，如果當前操作符小于等于則進行下面操作if (operstack.priority(ch) <= operstack.priority(operstack.peekTop())){num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);//把運算的結果加入棧numstack.push(res);//然后將當前操作符入符號棧operstack.push(ch);}else {operstack.push(ch);}}else {//為空直接入棧operstack.push(ch);}}else {//如果是數直接入棧(這里的1是字符'1'，要轉成數字1，ascll碼減48)numstack.push(ch - 48);}//讓index + 1，判斷是否掃描到expression最后index ++;if (index >= expression.length()){break;}}//掃描表達式完畢后，就順序的從 數棧和符號棧中pop出相應的數和符合，并運算while (true){//如果符號棧為空，則計算得到最后的結果，數棧中只有一個數字if (operstack.isEmpty()){break;}num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);numstack.push(res);}//打印結果(就是數棧中的最后一個數)System.out.printf("表達式 %s = %d",expression,numstack.pop());} }class ArrayStack1{private int maxSize; //棧大小private int[] stack;//數組模擬棧，數據放入數組里private int top = -1; //top表示棧頂，初始化為-1//構造器public ArrayStack1(int maxSize) {this.maxSize = maxSize;stack = new int[this.maxSize];}//返回棧頂的值，不出棧public int peekTop(){return stack[top];}//棧滿public boolean isFull(){return top == maxSize - 1;}//棧空public boolean isEmpty(){return top == -1;}// 入棧 pushpublic void push(int value){//判斷棧是否滿if (isFull()){System.out.println("棧滿，不能入棧");return;}top++;stack[top] = value;}//出棧public int pop(){if (isEmpty()){//拋出異常throw new RuntimeException("棧空");}int value = stack[top];top--;return value;}//遍歷棧(從棧頂開始)public void list(){if (isEmpty()){System.out.println("棧空");}for (int i = top; i>=0; i--){System.out.printf("stack[%d] = %d\n",i,stack[i]);}}//返回運算符優先級，優先級有程序員來確定，優先級使用數字表示//數字越大，優先級越高public int priority(int oper){if (oper == '*' || oper == '/'){return 1;}else if(oper == '+' || oper == '-'){return 0;}else {return -1;//假定目前的表達式只有 +-*/}}//判斷是否是一個運算符public boolean isOper(char val){return val == '+' || val == '-' || val == '*' || val == '/';}//計算方法public int cal(int num1, int num2, int oper){int res = 0; //計算結果switch (oper){case '+':res = num1 + num2;break;case '-':res = num2 - num1; //注意順序break;case '*':res = num1 * num2;break;case '/':res = num2 / num1;break;}return res;} }

找到是哪個地方有問題？，發現是一個數字就直接入棧了，那么需要在入數棧時，需要向后看，如果是數就拼接并繼續掃描，如果是符合才入棧。（那么就需要一個字符串變量用于拼接）

代碼

package stack;public class Calculator {public static void main(String[] args) {//完成表達式運算String expression = "71+2*6-2*2";//創建兩個棧，一個數棧，一個符號棧ArrayStack1 numstack = new ArrayStack1(10);ArrayStack1 operstack = new ArrayStack1(10);//定義需要的相關變量int index = 0; //用于掃描int num1 = 0, num2 = 0;int oper = 0;int res = 0;char ch = ' ';//將每次掃描得到的char保存給chString keepnum = ""; //用于拼接多位數//開始while循環的掃描expressionwhile(true){//一次得到expression 的每個字符ch = expression.substring(index,index+1).charAt(0);//判斷ch是字符還是數字if (operstack.isOper(ch)){//判斷符號棧是否為空if (!operstack.isEmpty()){//如果不為空，比較，如果當前操作符小于等于則進行下面操作if (operstack.priority(ch) <= operstack.priority(operstack.peekTop())){num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);//把運算的結果加入棧numstack.push(res);//然后將當前操作符入符號棧operstack.push(ch);}else {operstack.push(ch);}}else {//為空直接入棧operstack.push(ch);}}else {//如果是數直接入棧(這里的1是字符'1'，要轉成數字1，ascll碼減48)//如果有多位數，則需要改進代碼，不能直接入棧//思路：// 1.掃描到數字，需要繼續向后掃描，如果是數就拼接，繼續掃描，直到掃描是字符keepnum += ch;//如果ch是最后一位了則直接入棧if (index == expression.length()-1){numstack.push(Integer.parseInt(keepnum));}else {//判斷下一個字符是不是數字，如果是繼續掃描if (operstack.isOper(expression.substring(index + 1, index + 2).charAt(0))) {numstack.push(Integer.parseInt(keepnum));//這里一定要清空keepnumkeepnum = "";}}}//讓index + 1，判斷是否掃描到expression最后index ++;if (index >= expression.length()){break;}}//掃描表達式完畢后，就順序的從 數棧和符號棧中pop出相應的數和符合，并運算while (true){//如果符號棧為空，則計算得到最后的結果，數棧中只有一個數字if (operstack.isEmpty()){break;}num1 = numstack.pop();num2 = numstack.pop();oper = operstack.pop();res = numstack.cal(num1,num2,oper);numstack.push(res);}//打印結果(就是數棧中的最后一個數)System.out.printf("表達式 %s = %d",expression,numstack.pop());} }class ArrayStack1{private int maxSize; //棧大小private int[] stack;//數組模擬棧，數據放入數組里private int top = -1; //top表示棧頂，初始化為-1//構造器public ArrayStack1(int maxSize) {this.maxSize = maxSize;stack = new int[this.maxSize];}//返回棧頂的值，不出棧public int peekTop(){return stack[top];}//棧滿public boolean isFull(){return top == maxSize - 1;}//棧空public boolean isEmpty(){return top == -1;}// 入棧 pushpublic void push(int value){//判斷棧是否滿if (isFull()){System.out.println("棧滿，不能入棧");return;}top++;stack[top] = value;}//出棧public int pop(){if (isEmpty()){//拋出異常throw new RuntimeException("棧空");}int value = stack[top];top--;return value;}//遍歷棧(從棧頂開始)public void list(){if (isEmpty()){System.out.println("棧空");}for (int i = top; i>=0; i--){System.out.printf("stack[%d] = %d\n",i,stack[i]);}}//返回運算符優先級，優先級有程序員來確定，優先級使用數字表示//數字越大，優先級越高public int priority(int oper){if (oper == '*' || oper == '/'){return 1;}else if(oper == '+' || oper == '-'){return 0;}else {return -1;//假定目前的表達式只有 +-*/}}//判斷是否是一個運算符public boolean isOper(char val){return val == '+' || val == '-' || val == '*' || val == '/';}//計算方法public int cal(int num1, int num2, int oper){int res = 0; //計算結果switch (oper){case '+':res = num1 + num2;break;case '-':res = num2 - num1; //注意順序break;case '*':res = num1 * num2;break;case '/':res = num2 / num1;break;}return res;} }

總結

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