还有一个选择:https://github.com/stefanhaustein/expressionparser

我已经实现了一个简单而灵活的选项，以允许两者:

即时处理(Calculator.java, SetDemo.java) 构建和处理解析树(TreeBuilder.java)

上面链接的TreeBuilder是进行符号推导的CAS演示包的一部分。还有一个BASIC解释器的例子，我已经开始使用它来构建一个TypeScript解释器。

如果我们要实现它，那么我们可以使用下面的算法

While there are still tokens to be read in, 1.1 Get the next token. 1.2 If the token is: 1.2.1 A number: push it onto the value stack. 1.2.2 A variable: get its value, and push onto the value stack. 1.2.3 A left parenthesis: push it onto the operator stack. 1.2.4 A right parenthesis: 1 While the thing on top of the operator stack is not a left parenthesis, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. 2 Pop the left parenthesis from the operator stack, and discard it. 1.2.5 An operator (call it thisOp): 1 While the operator stack is not empty, and the top thing on the operator stack has the same or greater precedence as thisOp, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. 2 Push thisOp onto the operator stack. While the operator stack is not empty, 1 Pop the operator from the operator stack. 2 Pop the value stack twice, getting two operands. 3 Apply the operator to the operands, in the correct order. 4 Push the result onto the value stack. At this point the operator stack should be empty, and the value stack should have only one value in it, which is the final result.

我想无论你用什么方法做这个都会涉及到很多条件命题。但是对于单个操作，比如在你的例子中，你可以将它限制为4个if语句

String math = "1+4";

if (math.split("+").length == 2) {
    //do calculation
} else if (math.split("-").length == 2) {
    //do calculation
} ...

当你想要处理像“4+5*6”这样的多个操作时，它会变得更加复杂。

如果你试图构建一个计算器，那么我建议分别传递计算的每个部分(每个数字或运算符)，而不是作为一个单一的字符串。

在我的大学项目中，我正在寻找一个既支持基本公式又支持更复杂方程(特别是迭代运算符)的解析器/求值器。我发现了一个非常好的JAVA和。net开源库，叫做mXparser。我将给出几个例子，让大家对语法有一些感觉，如需进一步指导，请访问项目网站(特别是教程部分)。

https://mathparser.org/

https://mathparser.org/mxparser-tutorial/

https://mathparser.org/api/

举几个例子

一个简单的开始

Expression e = new Expression("( 2 + 3/4 + sin(pi) )/2");
double v = e.calculate()

2 -用户定义的参数和常量

Argument x = new Argument("x = 10");
Constant a = new Constant("a = pi^2");
Expression e = new Expression("cos(a*x)", x, a);
double v = e.calculate()

3 -用户定义的函数

Function f = new Function("f(x, y, z) = sin(x) + cos(y*z)");
Expression e = new Expression("f(3,2,5)", f);
double v = e.calculate()

4 -迭代

Expression e = new Expression("sum( i, 1, 100, sin(i) )");
double v = e.calculate()

最近发现的-如果你想尝试语法(并查看高级用例)，你可以下载由mXparser支持的标量计算器应用程序。

还有一个选择:https://github.com/stefanhaustein/expressionparser

我已经实现了一个简单而灵活的选项，以允许两者:

即时处理(Calculator.java, SetDemo.java) 构建和处理解析树(TreeBuilder.java)

上面链接的TreeBuilder是进行符号推导的CAS演示包的一部分。还有一个BASIC解释器的例子，我已经开始使用它来构建一个TypeScript解释器。

使用带有代码注入处理的JDK1.6 Javascript引擎尝试下面的示例代码。

import javax.script.ScriptEngine;
import javax.script.ScriptEngineManager;

public class EvalUtil {
private static ScriptEngine engine = new ScriptEngineManager().getEngineByName("JavaScript");
public static void main(String[] args) {
    try {
        System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || 5 >3 "));
        System.out.println((new EvalUtil()).eval("(((5+5)/2) > 5) || true"));
    } catch (Exception e) {
        e.printStackTrace();
    }
}
public Object eval(String input) throws Exception{
    try {
        if(input.matches(".*[a-zA-Z;~`#$_{}\\[\\]:\\\\;\"',\\.\\?]+.*")) {
            throw new Exception("Invalid expression : " + input );
        }
        return engine.eval(input);
    } catch (Exception e) {
        e.printStackTrace();
        throw e;
    }
 }
}