java计算数学表达式详解编程语言

  
import java.util.EmptyStackException; 
import java.util.Stack; 
   
public class CaculateFunction { 
    private static String[] TrnsInToSufix(String IFX)// PFX放后缀表达式，IFX为中缀表达式 
    { 
        String PFX[] = new String[IFX.length()]; 
        StringBuffer numBuffer = new StringBuffer();// 用来保存一个数的 
        Stack<String> s = new Stack<String>();// 放操作符 
        String a; 
        s.push("=");// 第一个为等号 
        int i = 0, j = 0; 
        char ch; 
        for (i = 0; i < IFX.length();) { 
            ch = IFX.charAt(i); 
            switch (ch) { 
            case '0': 
            case '1': 
            case '2': 
            case '3': 
            case '4': 
            case '5': 
            case '6': 
            case '7': 
            case '8': 
            case '9': 
                while (Character.isDigit(ch) || ch == '.')// 拼数 
                { 
                    numBuffer.append(ch); // 追加字符 
                    ch = IFX.charAt(++i); 
                } 
                PFX[j++] = numBuffer.toString();// break; 
                numBuffer = new StringBuffer(); // 清空已获取的运算数字 
                continue; // 这里要重新循环，因为i已经增加过了 
            case '(': 
                s.push("("); 
                break; 
            case ')': 
                while (s.peek() != "(") 
                    PFX[j++] = s.pop(); 
                break; 
            case '+': 
            case '-': 
                while (s.size() > 1 && s.peek() != "(") 
                    PFX[j++] = s.pop(); 
                a = String.valueOf(ch); 
                s.push(a); 
                break; 
            case '*': 
            case '/': 
                while (s.size() > 1 && (s.peek() == "*") || s.peek() == "/" 
                        || s.peek() == "s" || s.peek() == "c" 
                        || s.peek() == "t" || s.peek() == "^" 
                        || s.peek() == "√") 
                    // 优先级比较，与栈顶比较， 
                    PFX[j++] = s.pop();// 当前操作符优先级大于等于栈顶的弹出栈顶 
                a = String.valueOf(ch); 
                s.push(a); 
                break; 
            case 's': 
            case 'c': 
            case 't':// 三角函数 
                while (s.size() > 1 
                        && (s.peek() == "s" || s.peek() == "c" 
                                || s.peek() == "t" || s.peek() == "^" || s 
                                .peek() == "√")) 
                    // 优先级比较，与栈顶，大于等于的弹出 
                    PFX[j++] = s.pop(); 
                a = String.valueOf(ch); 
                s.push(a); 
                break; 
            case '^':// 幂 
            case '√':// 开方 
                while (s.size() > 1 && (s.peek() == "^" || s.peek() == "√")) 
                    PFX[j++] = s.pop(); 
                a = String.valueOf(ch); 
                s.push(a); 
                break; 
            } 
            i++; 
        } 
        while (s.size() > 1) 
            PFX[j++] = s.pop(); 
        PFX[j] = "="; 
   
        return PFX; 
    } 
   
    public static String Evaluate(String IFX)// 后缀表达式求值 
    { 
        String PFX[] = null; 
        try { 
            PFX = TrnsInToSufix(IFX); 
        } catch (EmptyStackException e) { 
            return "syntax error"; 
        } 
        int i = 0; 
        double x1, x2, n; 
        String str; 
        Stack<String> s = new Stack<String>(); 
        while (PFX[i] != "=") { 
            str = PFX[i]; 
            switch (str.charAt(0)) { 
            case '0': 
            case '1': 
            case '2': 
            case '3': 
            case '4': 
            case '5': 
            case '6': 
            case '7': 
            case '8': 
            case '9': 
                s.push(str); 
                break; 
            case '+': 
                x1 = Double.parseDouble(s.pop()); 
                x2 = Double.parseDouble(s.pop()); 
                n = x1 + x2; 
                s.push(String.valueOf(n)); 
                break; 
            case '-': 
                x1 = Double.parseDouble(s.pop()); 
                x2 = Double.parseDouble(s.pop()); 
                n = x2 - x1; 
                s.push(String.valueOf(n)); 
                break; 
            case '*': 
                x1 = Double.parseDouble(s.pop()); 
                x2 = Double.parseDouble(s.pop()); 
                n = x1 * x2; 
                s.push(String.valueOf(n)); 
                break; 
            case '/': 
                x1 = Double.parseDouble(s.pop()); 
                x2 = Double.parseDouble(s.pop()); 
                n = x2 / x1; 
                s.push(String.valueOf(n)); 
                break; 
            case 's': 
                x1 = Double.parseDouble(s.pop()); 
                n = Math.sin(x1 * Math.PI / 180); 
                s.push(String.valueOf(n)); 
                break; 
            case 'c': 
                x1 = Double.parseDouble(s.pop()); 
                n = Math.cos(x1 * Math.PI / 180); 
                s.push(String.valueOf(n)); 
                break; 
            case 't': 
                x1 = Double.parseDouble(s.pop()); 
                n = Math.tan(x1 * Math.PI / 180); 
                s.push(String.valueOf(n)); 
                break; 
            case '√': 
                x1 = Double.parseDouble(s.pop()); 
                n = Math.sqrt(x1); 
                s.push(String.valueOf(n)); 
                break;// 开方 
            case '^': 
                x1 = Double.parseDouble(s.pop()); 
                x2 = Double.parseDouble(s.pop()); 
                n = Math.pow(x2, x1); 
                s.push(String.valueOf(n)); 
                break; 
            } 
            i++; 
        } 
        String result = s.pop(); 
        return result; 
    } 
   
    public static void main(String args[]) { 
        System.out.println(Evaluate("(31 + 21) * 51 - (21 + 33) / 2 = ")); 
    } 
}