java中实现四则运算代码

最近上网查了一下,竟然没有找到用Java编写的四则运算的代码,就小写了一下.如有问题请大家反馈.

1.说明

代码只是实现了简单的四则运算,支持+,-,*,/,(,) 只能计算出正确的表达式的值,没有对非法表达式进行校验.

2.实现方法

第一步:将输入的字符串转换为List,主要是用来将String转换为原子:数值/运算符/括号

public List transStr(String str)
 {
 List strList = new ArrayList();
 
 
 String tmp = str.replaceAll("\d*", "");
 
 String curLet = null;
 
 int loc = 0;
 
 int len = tmp.length();
 for (int i = 0; i < len; i++)
 {
  curLet = tmp.substring(i, i + 1);
  loc = str.indexOf(curLet);
  
  if (!"".equals(str.substring(0, loc).trim()))
  {
  strList.add(str.substring(0, loc).trim());
  }
  strList.add(str.substring(loc, loc + 1));
  str = str.substring(loc + 1);
 }
 if (0 < str.length())
 {
  strList.add(str.trim());
 }
 return strList;
 }

第二步: 将原来的中缀表达式转换为后缀表达式,在四则运算中,后缀表达式是最方便计算的

public String[] midToEnd(List midList)
 {
 Stack embl = new Stack();
 Stack result = new Stack();
 
 Iterator it = midList.iterator();
 String curStr = null;
 while (it.hasNext())
 {
  curStr = (String) it.next();
  
  
  if(sign.containsKey(curStr))
  {
  
  if (0 == embl.size() || "(".equals(curStr))
  {
   embl.push(curStr);
  }
  else
  {
   
   if(")".equals(curStr))
   {
   while(!"(".equals((String)embl.peek()))
   {
    if(0 >= embl.size())
    {
    return null;
    }
    result.push(embl.pop());
   }
   embl.pop();
   }
   else
   {
   int p1 = Integer.parseInt((String) sign.get(curStr));
   int p2 = Integer.parseInt((String) sign.get(embl.peek()));
   
   
   if (p1 > p2)
   {
    embl.push(curStr);
   }
   else
   {
    while (p1 <= p2 || embl.size() > 0)
    {
    result.push(embl.pop());
    if(0 == embl.size())
    {
     break;
    }
    p2 = Integer.parseInt((String) sign.get(embl.peek()));
    }
    embl.push(curStr);
   }
   }
  }
  }
  else
  {
  result.push(curStr);
  }
 }
 
 while (0 < embl.size())
 {
  result.push(embl.pop());
 }
 
 int len = result.size();
 String[] ret = new String[len];
 for (int i = 0; i < len; i++)
 {
  ret[len - i - 1] = (String) result.pop();
 }
 
 return ret;
 }

第三步:将解析后缀表达式,返回计算的最终结果

 public Object calculate(String[] endStr)
 {
 int len = endStr.length;
 Stack calc = new Stack();
 double p2;
 double p1;
 for (int i = 0; i < len; i++)
 {
  if (sign.containsKey(endStr[i]))
  {
  try
  {
   p2 = Double.parseDouble((String) calc.pop());
   p1 = Double.parseDouble((String) calc.pop());
   calc.push(String.valueOf(simpleCalc(p1, p2,endStr[i])));
  }
  catch(NumberFormatException ex)
  {
   ex.printStackTrace();
   return "Input Error";
  }
  catch(Exception ex)
  {
   ex.printStackTrace();
   return "Input Error";
  }
  }
  else
  {
  calc.push(endStr[i]);
  }
 }
 
 if (1 == calc.size())
 {
  return calc.pop();
 }
 else
 {
  return "Input Error";
 }
 }
 
 
 public double simpleCalc(double p1, double p2, String oper)
 {
 
 switch(oper.charAt(0))
 {
   case '+':
     return p1 + p2;
   case '-':
     return p1 - p2;
   case '*':
     return p1 * p2;
   case '/':
     return p1 / p2;
   default:
    return p1;
 }
 }

第四步:运算符的优先级放在了缓存中进行提取

 private static HashMap sign = new HashMap();
 
 public CalculateExp()
 {
 sign.put(")", "3");
 sign.put("*", "2");
 sign.put("/", "2");
 sign.put("+", "1");
 sign.put("-", "1");
 sign.put("(", "0");
 }

完整代码

import java.util.ArrayList;
import java.util.HashMap;
import java.util.Iterator;
import java.util.List;
import java.util.Stack;


public class CalculateExp
{
	private static HashMap sign = new HashMap();
	
	public CalculateExp()
	{
		sign.put(")", "3");
		sign.put("*", "2");
		sign.put("/", "2");
		sign.put("+", "1");
		sign.put("-", "1");
		sign.put("(", "0");
	}
	
	public List transStr(String str)
	{
		List strList = new ArrayList();
		
		
		String tmp = str.replaceAll("\d*", "");
		
		String curLet = null;
		
		int loc = 0;
		
		int len = tmp.length();
		for (int i = 0; i < len; i++)
		{
			curLet = tmp.substring(i, i + 1);
			loc = str.indexOf(curLet);
			

			if (!"".equals(str.substring(0, loc).trim()))
			{
				strList.add(str.substring(0, loc).trim());
			}
			strList.add(str.substring(loc, loc + 1));
			str = str.substring(loc + 1);
		}
		if (0 < str.length())
		{
			strList.add(str.trim());
		}
		return strList;
	}

	
	public String[] midToEnd(List midList)
	{
		Stack embl = new Stack();
		Stack result = new Stack();
		
		Iterator it = midList.iterator();
		String curStr = null;
		while (it.hasNext())
		{
			curStr = (String) it.next();
			
			
			if(sign.containsKey(curStr))
			{
				
				if (0 == embl.size() || "(".equals(curStr))
				{
					embl.push(curStr);
				}
				else
				{
					
					if(")".equals(curStr))
					{
						while(!"(".equals((String)embl.peek()))
						{
							if(0 >= embl.size())
							{
								return null;
							}
							result.push(embl.pop());
						}
						embl.pop();
					}
					else
					{
						int p1 = Integer.parseInt((String) sign.get(curStr));
						int p2 = Integer.parseInt((String) sign.get(embl.peek()));
						
						
						if (p1 > p2)
						{
							embl.push(curStr);
						}
						else
						{
							while (p1 <= p2 || embl.size() > 0)
							{
								result.push(embl.pop());
								if(0 == embl.size())
								{
									break;
								}
								p2 = Integer.parseInt((String) sign.get(embl.peek()));
							}
							embl.push(curStr);
						}
					}
				}
			}
			else
			{
				result.push(curStr);
			}
		}
		
		while (0 < embl.size())
		{
			result.push(embl.pop());
		}
		
		int len = result.size();
		String[] ret = new String[len];
		for (int i = 0; i < len; i++)
		{
			ret[len - i - 1] = (String) result.pop();
		}
		
		return ret;
	}
	
	
	public Object calculate(String[] endStr)
	{
		int len = endStr.length;
		Stack calc = new Stack();
		double p2;
		double p1;
		for (int i = 0; i < len; i++)
		{
			if (sign.containsKey(endStr[i]))
			{
				try
				{
					p2 = Double.parseDouble((String) calc.pop());
					p1 = Double.parseDouble((String) calc.pop());
					calc.push(String.valueOf(simpleCalc(p1, p2,endStr[i])));
				}
				catch(NumberFormatException ex)
				{
					ex.printStackTrace();
					return "Input Error";
				}
				catch(Exception ex)
				{
					ex.printStackTrace();
					return "Input Error";
				}
			}
			else
			{
				calc.push(endStr[i]);
			}
		}
		
		if (1 == calc.size())
		{
			return calc.pop();
		}
		else
		{
			return "Input Error";
		}
	}
	
	
	public double simpleCalc(double p1, double p2, String oper)
	{
		
		switch(oper.charAt(0))
		{
		  case '+':
		    return p1 + p2;
		  case '-':
		    return p1 - p2;
		  case '*':
		    return p1 * p2;
		  case '/':
		    return p1 / p2;
		  default:
		  	return p1;
		}
	}
	
	
	public static void main(String[] args)
	{
		CalculateExp ce = new CalculateExp();
		String tmp = "3+12+25*(20-20/4+10";
		String ret = (String) ce.calculate(ce.midToEnd(ce
				.transStr(tmp)));
		double value = 0;
    try
    {
    	value = Double.parseDouble(ret);
    }
    catch (NumberFormatException ex)
    {
    	System.out.print(ret);
    }
		System.out.print(value);
	}
}

以下是其他网友的补充

代码的思路是通过正则判断计算每个最小的计算单元。以下是代码：

import java.math.BigDecimal;
import java.util.regex.Matcher;
import java.util.regex.Pattern;


public class CalculatorUtil {

 public static BigDecimal arithmetic(String exp){
  if(!exp.matches("\d+")){
   String result = parseExp(exp).replaceAll("[\[\]]", "");
   return new BigDecimal(result);
  }else{
   return new BigDecimal(exp);
  }
 }
 
 private static String minExp="^((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\]))[\+\-\*\/]((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\]))$";
 
 private static String noParentheses="^[^\(\)]+$";
 
 private static String priorOperatorExp="(((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\]))[\*\/]((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\])))";
 
 private static String operatorExp="(((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\]))[\+\-]((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\])))";
 
 private static String minParentheses="\([^\(\)]+\)";
 
 
 private static String parseExp(String expression){
  //方法进入 先替换空格，在去除运算两边的()号
  expression=expression.replaceAll("\s+", "").replaceAll("^\(([^\(\)]+)\)$", "$1");
  
  //最小表达式计算
  if(expression.matches(minExp)){
   String result=calculate(expression);
   return Double.parseDouble(result)>=0?result:"["+result+"]";
  }
  //计算不带括号的四则运算
  if(expression.matches(noParentheses)){
   Pattern patt=Pattern.compile(priorOperatorExp);
   Matcher mat=patt.matcher(expression);
   if(mat.find()){
    String tempMinExp=mat.group();
    expression=expression.replaceFirst(priorOperatorExp, parseExp(tempMinExp));
   }else{
    patt=Pattern.compile(operatorExp);
    mat=patt.matcher(expression);
    
    if(mat.find()){
     String tempMinExp=mat.group();
     expression=expression.replaceFirst(operatorExp, parseExp(tempMinExp));
    }
   }
   return parseExp(expression);
  }
  
  //计算带括号的四则运算
  Pattern patt=Pattern.compile(minParentheses);
  Matcher mat=patt.matcher(expression);
  if(mat.find()){
   String tempMinExp=mat.group();
   expression=expression.replaceFirst(minParentheses, parseExp(tempMinExp));
  }
  return parseExp(expression);
 }
 
 private static String calculate(String exp){
  exp=exp.replaceAll("[\[\]]", "");
  String number[]=exp.replaceFirst("(\d)[\+\-\*\/]", "$1,").split(",");
  BigDecimal number1=new BigDecimal(number[0]);
  BigDecimal number2=new BigDecimal(number[1]);
  BigDecimal result=null;
  
  String operator=exp.replaceFirst("^.*\d([\+\-\*\/]).+$", "$1");
  if("+".equals(operator)){
   result=number1.add(number2);
  }else if("-".equals(operator)){
   result=number1.subtract(number2);
  }else if("*".equals(operator)){
   result=number1.multiply(number2);
  }else if("/".equals(operator)){
   //第二个参数为精度，第三个为四色五入的模式
   result=number1.divide(number2,5,BigDecimal.ROUND_CEILING);
  }
  
  return result!=null?result.toString():null;
 }
 
}

代码原本是一个博客，原来代码没有注释而且存在BUG，我稍微修稿了一哈添加了注释。在这里做个笔记，方便以后用

另为表示对原作者的敬意，附上原始代码

public class Arithmetic {
 public static void main(String args[]){
 System.out.println(arithmetic("2.2+((3+4)*2-22)/2*3.2"));
 }
 public static double arithmetic(String exp){
 String result = parseExp(exp).replaceAll("[\[\]]", "");
 return Double.parseDouble(result);
 }
 
 public static String parseExp(String expression){
 //String numberReg="^((?!0)\d+(\.\d+(?=0?result:"["+result+"]";
 }
 //计算不带括号的四则运算
 String noParentheses="^[^\(\)]+$";
 String priorOperatorExp="(((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\]))[\*\/]((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\])))";
 String operatorExp="(((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\]))[\+\-]((\d+(\.\d+)?)|(\[\-\d+(\.\d+)?\])))";
 if(expression.matches(noParentheses)){
 Pattern patt=Pattern.compile(priorOperatorExp);
 Matcher mat=patt.matcher(expression);
 if(mat.find()){
 String tempMinExp=mat.group();
 expression=expression.replaceFirst(priorOperatorExp, parseExp(tempMinExp));
 }else{
 patt=Pattern.compile(operatorExp);
 mat=patt.matcher(expression);
 
 if(mat.find()){
  String tempMinExp=mat.group();
  expression=expression.replaceFirst(operatorExp, parseExp(tempMinExp));
 }
 }
 return parseExp(expression);
 }
 //计算带括号的四则运算
 String minParentheses="\([^\(\)]+\)";
 Pattern patt=Pattern.compile(minParentheses);
 Matcher mat=patt.matcher(expression);
 if(mat.find()){
 String tempMinExp=mat.group();
 expression=expression.replaceFirst(minParentheses, parseExp(tempMinExp));
 }
 return parseExp(expression);
 }
 
 public static String calculate(String exp){
 exp=exp.replaceAll("[\[\]]", "");
 String number[]=exp.replaceFirst("(\d)[\+\-\*\/]", "$1,").split(",");
 BigDecimal number1=new BigDecimal(number[0]);
 BigDecimal number2=new BigDecimal(number[1]);
 BigDecimal result=null;
 
 String operator=exp.replaceFirst("^.*\d([\+\-\*\/]).+$", "$1");
 if("+".equals(operator)){
 result=number1.add(number2);
 }else if("-".equals(operator)){
 result=number1.subtract(number2);
 }else if("*".equals(operator)){
 result=number1.multiply(number2);
 }else if("/".equals(operator)){
 result=number1.divide(number2);
 }
 
 return result!=null?result.toString():null;
 }
}

最后给大家分享一个网友的实现方法，个人感觉也很不错

import java.util.Stack; 
  
public class Operate {  
 private Stack priStack = new Stack();// 操作符栈  
 private Stack numStack = new Stack();;// 操作数栈  
  
   
 public int caculate(String str) {  
  // 1.判断string当中有没有非法字符  
  String temp;// 用来临时存放读取的字符  
  // 2.循环开始解析字符串，当字符串解析完，且符号栈为空时，则计算完成  
  StringBuffer tempNum = new StringBuffer();// 用来临时存放数字字符串(当为多位数时)  
  StringBuffer string = new StringBuffer().append(str);// 用来保存，提高效率  
  
  while (string.length() != 0) {  
   temp = string.substring(0, 1);  
   string.delete(0, 1);  
   // 判断temp，当temp为操作符时  
   if (!isNum(temp)) {  
    // 1.此时的tempNum内即为需要操作的数，取出数，压栈，并且清空tempNum  
    if (!"".equals(tempNum.toString())) {  
     // 当表达式的第一个符号为括号  
     int num = Integer.parseInt(tempNum.toString());  
     numStack.push(num); 
     tempNum.delete(0, tempNum.length());  
    }  
    // 用当前取得的运算符与栈顶运算符比较优先级：若高于，则因为会先运算，放入栈顶；若等于，因为出现在后面，所以会后计算，所以栈顶元素出栈，取出操作数运算；  
    // 若小于，则同理，取出栈顶元素运算，将结果入操作数栈。  
  
    // 判断当前运算符与栈顶元素优先级，取出元素，进行计算(因为优先级可能小于栈顶元素，还小于第二个元素等等，需要用循环判断)  
    while (!compare(temp.charAt(0)) && (!priStack.empty())) { 
     int a = (int) numStack.pop();// 第二个运算数  
     int b = (int) numStack.pop();// 第一个运算数  
     char ope = priStack.pop();  
     int result = 0;// 运算结果  
     switch (ope) {  
     // 如果是加号或者减号，则  
     case '+':  
      result = b + a;  
      // 将操作结果放入操作数栈  
      numStack.push(result);  
      break;  
     case '-':  
      result = b - a;  
      // 将操作结果放入操作数栈  
      numStack.push(result);  
      break;  
     case '*':  
      result = b * a;  
      // 将操作结果放入操作数栈  
      numStack.push(result);  
      break;  
     case '/':  
      result = b / a;// 将操作结果放入操作数栈  
      numStack.push(result);  
      break;  
     }  
  
    }  
    // 判断当前运算符与栈顶元素优先级， 如果高，或者低于平，计算完后，将当前操作符号，放入操作符栈  
    if (temp.charAt(0) != '#') {  
     priStack.push(new Character(temp.charAt(0)));  
     if (temp.charAt(0) == ')') {// 当栈顶为'('，而当前元素为')'时，则是括号内以算完，去掉括号  
      priStack.pop();  
      priStack.pop();  
     }  
    }  
   } else  
    // 当为非操作符时（数字）  
    tempNum = tempNum.append(temp);// 将读到的这一位数接到以读出的数后(当不是个位数的时候)  
  }  
  return numStack.pop();  
 }  
  
   
 private boolean isNum(String temp) {  
  return temp.matches("[0-9]");  
 }  
  
   
 private boolean compare(char str) {  
  if (priStack.empty()) {  
   // 当为空时，显然 当前优先级最低，返回高  
   return true;  
  }  
  char last = (char) priStack.lastElement();  
  // 如果栈顶为'('显然，优先级最低，')'不可能为栈顶。  
  if (last == '(') {  
   return true;  
  }  
  switch (str) {  
  case '#':  
   return false;// 结束符  
  case '(':  
   // '('优先级最高,显然返回true  
   return true;  
  case ')':  
   // ')'优先级最低，  
   return false;  
  case '*': {  
   // '*/'优先级只比'+-'高  
   if (last == '+' || last == '-')  
    return true;  
   else  
    return false;  
  }  
  case '/': {  
   if (last == '+' || last == '-')  
    return true;  
   else  
    return false;  
  }  
   // '+-'为最低，一直返回false  
  case '+':  
   return false;  
  case '-':  
   return false;  
  }  
  return true;  
 }  
  
 public static void main(String args[]) {  
  Operate operate = new Operate();  
  int t = operate.caculate("(3+4*(4*10-10/2)#");  
  System.out.println(t);  
 }  
  
}