比较java中的字符串并删除它们相同的字符串部分

Question

I have two strings with me:

s1="MICROSOFT"
s2="APPLESOFT"

I need to compare the strings and remove the duplicate part (always towards the end) from the second string. So I should get "MICROSOFT" and "APPLE" as output.

I have compared both the strings character by character.

               String s1 = "MICROSOFT";
               String s2 = "APPLESOFT";

               for(int j=0; j<s1.length(); j++)
               {
                   char c1 = s1.charAt(j);
                   char c2 = s2.charAt(j);

                   if(c1==c2)
                       System.out.println("Match found!!!");
                   else
                       System.out.println("No match found!");
               }

It should check the strings and if the two strings have same characters until the end of string, then I need to remove that redundant part, SOFT in this case, from the second string. But I can't think of how to proceed from here.

There can be more duplicates...but we have to remove only those which are continuously identical. if i have APPWWSOFT and APPLESOFT, i should get APPLE again in the second string since we got LE different than WW in between

Can you guys please help me out here?

Answer 1

Search and read about Longest Common Subsequence , you can find efficient algorithms to find out the LCS of two input strings. After finding the LCS of the input strings, it is easy to manipulate the inputs. For example, in your case an LCS algorithm will find "SOFT" as the LCS of these two strings, then you might check whether the LCS is in the final part of the 2nd input and then remove it easily. I hope this idea helps.

An example LCS code in Java is here, try it: http://introcs.cs.princeton.edu/java/96optimization/LCS.java.html

Example scenario (pseudocode):

input1: "MISROSOFT";
input2: "APPLESOFT";

execute LCS(input1, input2);
store the result in lcs, now lcs = "SOFT";

iterate over the characters of input2,
if a character exists in lcs then remove it from input2.

Answer 2

As far as I understand, you want to remove any identical characters from the two strings. By identical I mean: same position and same character(code). I think the following linear complexity solution is the simplest:

 StringBuilder sb1 = new StringBuilder();
 StringBuilder sb2 = new StringBuilder(); //if you want to remove the identical char 
                                          //only from one string you don't need the 2nd sb
 char c;
 for(int i = 0; i<Math.min(s1.length,s2.length);i++){
     if((c = s1.charAt(i)) != s2.charAt(i)){
           sb1.append(c);
     }
 }
 return sb1.toString();

Answer 3

Try this algo- Create characters sequences of your first string and find it in second string.

performance -
Average case = (s1.length()-1)sq

public class SeqFind {
    public static String searchReplace(String s1,String s2) {
        String s3;
        boolean brk=false;
        for(int j=s1.length();j>0&&!brk;j--){
        for (int i = j-4; i > 0; i--) {
            String string = s1.substring( i,j);
            if(s2.contains(string)){
                System.out.println(s2+" - "+string+" "+s2.replace( string,""));
                brk=true;
                break;
            }
        }
    }
        return s3;      
    }
    public static void main(String[] args) {
        String s1 = "MICROSOFT";
        String s2 = "APPLESOFT";
        String s3 = searchReplace(s1,s2);
    }
}

Out put - APPLESOFT - SOFT - APPLE

Answer 4

You should rather use StringBuffer if you want your String to be modified..

And in this case, you can have one extra StringBuffer , in which you can keep on appending non-matching character: -

    StringBuffer s1 = new StringBuffer("MICROSOFT");
    StringBuffer s2 = new StringBuffer("APPLESOFT");
    StringBuffer s3 = new StringBuffer();

    for(int j=0; j<s1.length(); j++)
    {
        char c1 = s1.charAt(j);
        char c2 = s2.charAt(j);

        if(c1==c2) {
            System.out.println("Match found!!!");
        } else {
            System.out.println("No match found!");
            s3.append(c1);
        }
    }
    s1 = s3;
    System.out.println(s1);    // Prints "MICRO"

Answer 5

   public class Match {

public static void main(String[] args)
{
    String s1="MICROSOFT";
    String s2="APPLESOFT";
    String[] s=new String[10];
    String s3;
    int j=0,k=0;
    for(int i=s2.length();i>0;i--)
    {
        s[j]=s2.substring(k,s2.length());
        if(s1.contains(s[j]))
        {
            s3=s2.substring(0,j);
                                 System.out.println(s1+""+s3);

            System.exit(0);

        }
        else
        {
            System.out.println("");
        }
                                j++;
                                k++;
    }


}

     }

I have edited the code you can give it an another try.

Answer 6

try this, not tested thou

 String s1 = "MICROSOFT";
         String s2 = "APPLESOFT";
         String s3="";
         for(int j=0; j<s1.length(); j++)
         {
             if(s1.charAt(j)==s2.charAt(j)){
                 s3+=s1.charAt(j);
             }
         }
         System.out.println(s1.replace(s3, " ") + " \n"+ s2.replace(s3, " "));

Answer 7

I have solved my problem after racking some brains off. Please feel free to correct/improve/refine my code. The code not only works for "MICROSOFT" and "APPLESOFT" inputs, but also for inputs like "APPWWSOFT" and "APPLESOFT" (i needed to remove the continuous duplicates from the end - SOFT in both the above inputs). I'm in the learning stage and I'll appreciate any valuable inputs.

public class test
    {           
        public static void main(String[] args)
        {
            String s1 = "MICROSOFT";
            String s2 = "APPLESOFT";

            int counter1=0;
            int counter2=0;

            String[] test = new String[100];
            test[0]="";

            for(int j=0; j<s1.length(); j++)
            {
                char c1 = s1.charAt(j);
                char c2 = s2.charAt(j);

                if(c1==c2)
                {
                    if(counter1==counter2)
                    {
                        //System.out.println("Match found!!!");
                        test[0]=test[0]+c2;
                        counter2++;
                        //System.out.println("Counter 2: "+counter2);
                    }
                    else
                        test[0]="";
                }
               else
               {
                   //System.out.print("No match found!");
                   //System.out.println("Counter 2: "+counter2);
                   counter2=counter1+1;
                   test[0]="";
               }

               counter1++;
               //System.out.println("Counter 1: "+counter1);
                           }

             System.out.println(test[0]);
             System.out.println(s2.replaceAll(test[0]," "));
        }
    }