百行以内实现复杂数学表达式计算

kuoying

2020-05-12

一改以前

本次先上代码

package good;
//Evaluate complex expressions
import java.util.Scanner;

public class Example {

  public static void main(String[] arg) {//test

    Scanner in = new Scanner(System.in);
//    String one = in.next();
    try {
      System.out.println("(1.5+1)+(1+1)=" + RemoveBrackets(new StringBuffer("(1.5+1)+(1+1)")));
    }
    catch (Exception e){
      System.out.println("Please check if the expression is valid if there is an error");
    }
  }

  public static double RemoveBrackets(StringBuffer a) {//Bracket removal method

    while (a.toString().indexOf(‘(‘) != -1) {
      int i = a.toString().indexOf(‘(‘);
      int j = GetBracketPosition(a.toString(), i + 1);
      String to = a.toString().substring(i + 1, j);
      a.replace(i, j + 1, String.valueOf(RemoveBrackets(new StringBuffer(to))));
    }
    return OnlyMultiplication(a.toString());//计算除乘法以外的算式
  }


  //A method for calculating addition, subtraction, multiplication, and division without parentheses
  private static double OnlyMultiplication(String a) {//To eliminate the formula other than multiplication

    char[] b = a.toCharArray();
    for (int i = 0; i < a.length(); i++) {/*I‘m going to check the addition and subtraction because I‘m going to multiply and divide and then I‘m going to add and subtract*/

      if (b[i] == ‘+‘) {
        return OnlyMultiplication(a.substring(0, i)) + OnlyMultiplication(a.substring(i + 1));//Prevent double parenthesis
      }

      if (b[i] == ‘-‘) {
        return OnlyMultiplication(a.substring(0, i)) - OnlyMultiplication(a.substring(i + 1));
      }
    }//So let‘s go over here and do the subtraction and the subtraction
    //That means there‘s no plus or minus sign
    //A separate multiplication or division
    String[] sum1 = a.split("/");//The second half of this method is to remove the division sign

    double too = Double.valueOf(CalculateMultiplication(sum1[0]));
    for (int i = 1; i < sum1.length; i++) {
      too = too * 1.0 / Double.valueOf(sum1[i]);
    }
    return too;
  }

  //So the only way to multiply is to multiply
  private static double CalculateMultiplication(String aaa) {//Remove the multiplication sign
    double to = 1.0;
    if (aaa.indexOf(‘*‘) == -1) {//Without a multiplication sign
      return Double.valueOf(aaa);
    }
    String[] too = aaa.split("\\*");

    for (int i = 0; i < too.length; i++) {
      to = to * Double.valueOf(too[i]);
    }
    return to;
  }

  private static int GetBracketPosition(String a, int b) {//Find the corresponding close parenthesis position
    int count1 = 1;
    int count2 = 0;
    char[] aa = a.toCharArray();
    for (int i = b; i < a.length(); i++) {
      if (aa[i] == ‘(‘) {//Prevent double parenthesis
        count1++;
      }
      if (aa[i] == ‘)‘) {
        count2++;
        if (count2 == count1) {
          return i;
        }
      }
    }
    return 0;
  }
}

下面是运行结果

百行以内实现复杂数学表达式计算