线性栈实现中缀表达式计算器
作者:互联网
线性栈实现中缀表达式计算器
方法代码:
以(10+20/2*3)/2+8为例(计算结果为28):
private static int evaluateExpression(String expression){
//创建两个栈,一个存放符号,一个存放数字
ArrayStack<Character> operatorStack = new ArrayStack<>();
ArrayStack<Integer> numberStack = new ArrayStack<>();
expression = insertBlanks(expression);//空格插入方法
String[] tokens = expression.split(" "); //按空格分割后用作遍历
for (String token : tokens) { //token ==tokens[i]
//过滤空串
if (token.length() == 0) {
continue;
} else if (token.equals("+") || token.equals("-")) {
while (!operatorStack.isEmpty() && (operatorStack.peek() == '+' || operatorStack.peek() == '-' || operatorStack.peek() == '*' || operatorStack.peek() == '/')) {
processAnOperator(numberStack, operatorStack);//弹栈计算方法
}
//如果符号栈为空或栈顶是(则直接将符号入栈
operatorStack.push(token.charAt(0));
} else if (token.equals("*") || token.equals("/")) {
while (!operatorStack.isEmpty() && (operatorStack.peek() == '*' || operatorStack.peek() == '/')) {
processAnOperator(numberStack, operatorStack);
}
operatorStack.push(token.charAt(0));
} else if (token.equals("(")) {
operatorStack.push(token.charAt(0));
} else if (token.equals(")")) {
while (operatorStack.peek() != '(') {
processAnOperator(numberStack, operatorStack);
}
operatorStack.pop();
} else { //遍历到数字
numberStack.push(new Integer(token));
}
}
while (!operatorStack.isEmpty()){
processAnOperator(numberStack, operatorStack);
}
return numberStack.pop();
}
空格插入和弹栈计算
总体思路:
空格插入方法
//遍历字符串,遇到非数字字符,在新字符串中加入空格c空格,遇到数字直接加入,
//而后以空格为分隔符分割就不会把例如10变成1,0
private static String insertBlanks(String expression){
StringBuilder sb = new StringBuilder();
for (int i = 0;i < expression.length();i++){
char c = expression.charAt(i);
if (c=='(' || c==')' || c=='+' || c=='-' || c=='*' || c=='/'){
sb.append(' ');
sb.append(c);
sb.append(' ');
}else {
sb.append(c);
}
}
return sb.toString();
}
}
弹栈计算方法:
//符号和数字弹栈计算,并将结果入栈
private static void processAnOperator(ArrayStack<Integer> numberStack, ArrayStack<Character> operatorStack){
char op = operatorStack.pop();
int num1 = numberStack.pop();
int num2 = numberStack.pop();
if (op == '+'){
numberStack.push(num2 + num1);
}else if (op == '-'){
numberStack.push(num2 - num1);
}else if (op == '*'){
numberStack.push(num2 * num1);
}else {
numberStack.push(num2 /num1);
}
}
总体源代码:
package p2.线性结构;
//中缀表达式计算器
public class InfixCalculator {
public static void main(String[] args) {
String experssion = "(10+20/2*3)/2+8";
int result = evaluateExpression(experssion);
System.out.println(result);
}
private static int evaluateExpression(String expression){
//创建两个栈,一个存放符号,一个存放数字
ArrayStack<Character> operatorStack = new ArrayStack<>();
ArrayStack<Integer> numberStack = new ArrayStack<>();
expression = insertBlanks(expression);
String[] tokens = expression.split(" "); //按空格分割后用作遍历
for (String token : tokens) { //token ==tokens[i]
//过滤空串
if (token.length() == 0) {
continue;
} else if (token.equals("+") || token.equals("-")) {
while (!operatorStack.isEmpty() && (operatorStack.peek() == '+' || operatorStack.peek() == '-' || operatorStack.peek() == '*' || operatorStack.peek() == '/')) {
processAnOperator(numberStack, operatorStack);
}
//如果符号栈为空或栈顶是(则直接将符号入栈
operatorStack.push(token.charAt(0));
} else if (token.equals("*") || token.equals("/")) {
while (!operatorStack.isEmpty() && (operatorStack.peek() == '*' || operatorStack.peek() == '/')) {
processAnOperator(numberStack, operatorStack);
}
operatorStack.push(token.charAt(0));
} else if (token.equals("(")) {
operatorStack.push(token.charAt(0));
} else if (token.equals(")")) {
while (operatorStack.peek() != '(') {
processAnOperator(numberStack, operatorStack);
}
operatorStack.pop();
} else { //遍历到数字
numberStack.push(new Integer(token));
}
}
while (!operatorStack.isEmpty()){
processAnOperator(numberStack, operatorStack);
}
return numberStack.pop();
}
//符号和数字弹栈计算,并将结果入栈
private static void processAnOperator(ArrayStack<Integer> numberStack, ArrayStack<Character> operatorStack){
char op = operatorStack.pop();
int num1 = numberStack.pop();
int num2 = numberStack.pop();
if (op == '+'){
numberStack.push(num2 + num1);
}else if (op == '-'){
numberStack.push(num2 - num1);
}else if (op == '*'){
numberStack.push(num2 * num1);
}else {
numberStack.push(num2 /num1);
}
}
//遍历字符串,遇到非数字字符,在新字符串中加入空格c空格,遇到数字直接加入,
//而后以空格为分隔符分割就不会把例如10变成1,0
private static String insertBlanks(String expression){
StringBuilder sb = new StringBuilder();
for (int i = 0;i < expression.length();i++){
char c = expression.charAt(i);
if (c=='(' || c==')' || c=='+' || c=='-' || c=='*' || c=='/'){
sb.append(' ');
sb.append(c);
sb.append(' ');
}else {
sb.append(c);
}
}
return sb.toString();
}
}
标签:peek,numberStack,operatorStack,else,token,计算器,push,表达式,中缀 来源: https://blog.csdn.net/weixin_54986763/article/details/122404059