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标题: java计算数学表达式（中缀转后缀加运算） [打印本页]

作者: sansu 时间: 2015-9-10 22:57
标题: java计算数学表达式（中缀转后缀加运算）

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 != "=") {
         str = PFX;
         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 = "));
}
}

作者: wu199309023 时间: 2015-9-10 22:59
好牛逼呀这么长的代码

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