可以使用Djikstra的分流码算法将中缀表示法中的任何表达式字符串转换为后缀表示法。然后，算法的结果可以作为后缀算法的输入，并返回表达式的结果。

我在这里写了一篇关于它的文章，用java实现

看来应该由JEP来做这项工作

像RHINO或NASHORN这样的外部库可以用来运行javascript。javascript可以计算简单的公式，而不需要对字符串进行分割。如果代码写得好，也不会对性能造成影响。 下面是一个使用RHINO -的示例

public class RhinoApp {
    private String simpleAdd = "(12+13+2-2)*2+(12+13+2-2)*2";

public void runJavaScript() {
    Context jsCx = Context.enter();
    Context.getCurrentContext().setOptimizationLevel(-1);
    ScriptableObject scope = jsCx.initStandardObjects();
    Object result = jsCx.evaluateString(scope, simpleAdd , "formula", 0, null);
    Context.exit();
    System.out.println(result);
}

对于JDK1.6，您可以使用内置的Javascript引擎。

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

public class Test {
  public static void main(String[] args) throws ScriptException {
    ScriptEngineManager mgr = new ScriptEngineManager();
    ScriptEngine engine = mgr.getEngineByName("JavaScript");
    String foo = "40+2";
    System.out.println(engine.eval(foo));
    } 
}

我写了算术表达式的eval方法来回答这个问题。它可以做加法、减法、乘法、除法、求幂(使用^符号)，以及一些基本函数，如平方根。它支持使用(…)进行分组，并获得正确的操作符优先级和结合规则。

public static double eval(final String str) {
    return new Object() {
        int pos = -1, ch;
        
        void nextChar() {
            ch = (++pos < str.length()) ? str.charAt(pos) : -1;
        }
        
        boolean eat(int charToEat) {
            while (ch == ' ') nextChar();
            if (ch == charToEat) {
                nextChar();
                return true;
            }
            return false;
        }
        
        double parse() {
            nextChar();
            double x = parseExpression();
            if (pos < str.length()) throw new RuntimeException("Unexpected: " + (char)ch);
            return x;
        }
        
        // Grammar:
        // expression = term | expression `+` term | expression `-` term
        // term = factor | term `*` factor | term `/` factor
        // factor = `+` factor | `-` factor | `(` expression `)` | number
        //        | functionName `(` expression `)` | functionName factor
        //        | factor `^` factor
        
        double parseExpression() {
            double x = parseTerm();
            for (;;) {
                if      (eat('+')) x += parseTerm(); // addition
                else if (eat('-')) x -= parseTerm(); // subtraction
                else return x;
            }
        }
        
        double parseTerm() {
            double x = parseFactor();
            for (;;) {
                if      (eat('*')) x *= parseFactor(); // multiplication
                else if (eat('/')) x /= parseFactor(); // division
                else return x;
            }
        }
        
        double parseFactor() {
            if (eat('+')) return +parseFactor(); // unary plus
            if (eat('-')) return -parseFactor(); // unary minus
            
            double x;
            int startPos = this.pos;
            if (eat('(')) { // parentheses
                x = parseExpression();
                if (!eat(')')) throw new RuntimeException("Missing ')'");
            } else if ((ch >= '0' && ch <= '9') || ch == '.') { // numbers
                while ((ch >= '0' && ch <= '9') || ch == '.') nextChar();
                x = Double.parseDouble(str.substring(startPos, this.pos));
            } else if (ch >= 'a' && ch <= 'z') { // functions
                while (ch >= 'a' && ch <= 'z') nextChar();
                String func = str.substring(startPos, this.pos);
                if (eat('(')) {
                    x = parseExpression();
                    if (!eat(')')) throw new RuntimeException("Missing ')' after argument to " + func);
                } else {
                    x = parseFactor();
                }
                if (func.equals("sqrt")) x = Math.sqrt(x);
                else if (func.equals("sin")) x = Math.sin(Math.toRadians(x));
                else if (func.equals("cos")) x = Math.cos(Math.toRadians(x));
                else if (func.equals("tan")) x = Math.tan(Math.toRadians(x));
                else throw new RuntimeException("Unknown function: " + func);
            } else {
                throw new RuntimeException("Unexpected: " + (char)ch);
            }
            
            if (eat('^')) x = Math.pow(x, parseFactor()); // exponentiation
            
            return x;
        }
    }.parse();
}

例子:

System.out.println(eval("((4 - 2^3 + 1) * -sqrt(3*3+4*4)) / 2"));

输出:7.5(正确)

该解析器是递归下降解析器，因此在内部为其语法中的每个操作符优先级使用单独的解析方法。我故意保持简短，但以下是一些你可能想要扩展的想法:

Variables: The bit of the parser that reads the names for functions can easily be changed to handle custom variables too, by looking up names in a variable table passed to the eval method, such as a Map<String,Double> variables. Separate compilation and evaluation: What if, having added support for variables, you wanted to evaluate the same expression millions of times with changed variables, without parsing it every time? It's possible. First define an interface to use to evaluate the precompiled expression: @FunctionalInterface interface Expression { double eval(); } Now to rework the original "eval" function into a "parse" function, change all the methods that return doubles, so instead they return an instance of that interface. Java 8's lambda syntax works well for this. Example of one of the changed methods: Expression parseExpression() { Expression x = parseTerm(); for (;;) { if (eat('+')) { // addition Expression a = x, b = parseTerm(); x = (() -> a.eval() + b.eval()); } else if (eat('-')) { // subtraction Expression a = x, b = parseTerm(); x = (() -> a.eval() - b.eval()); } else { return x; } } } That builds a recursive tree of Expression objects representing the compiled expression (an abstract syntax tree). Then you can compile it once and evaluate it repeatedly with different values: public static void main(String[] args) { Map<String,Double> variables = new HashMap<>(); Expression exp = parse("x^2 - x + 2", variables); for (double x = -20; x <= +20; x++) { variables.put("x", x); System.out.println(x + " => " + exp.eval()); } } Different datatypes: Instead of double, you could change the evaluator to use something more powerful like BigDecimal, or a class that implements complex numbers, or rational numbers (fractions). You could even use Object, allowing some mix of datatypes in expressions, just like a real programming language. :)

这个答案中的所有代码都已发布到公共领域。玩得开心!

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。