比较两个String数组，分别返回匹配的String和不匹配的String

Question

I am developing an android application. I am trying to compare two string arrays and return the matches , mismatches. I tried couple of solutions but it's not working at all. I need matches and mismatches separately.

Here are the arrays with example data

String[] number_one = { "info@fn.ca" , "+122637867"  , "486" , "smbr" , "9946567" };

String[] number_two = { "+122637867" , "486" , "nrkZone"  , "smbr'};

Here is the example code i am working on.

          for(int n =0; n < number_one.length; n++){

            if(number_one[n] == number_two[n]){
                 Log.d("Cursor" ,number_one[n]);
            }else{
                 Log.d("Cursor" ,number_two[n]);
             }

           }

I want outputs like (Example data)

 Matched : +122637867 , 486 , smbr
 Mismatched :  info@fn.ca , nrkZone , 9946567

Answer 1

I probably did more than I should have, and maybe even made it over complicated. But I'll let you guys decide that.

The first thing that I did was to use HashSets to store the matched and mismatched values. This is useful because they only store one of each value.

String[] number_one = { "info@fn.ca" , "+122637867"  , "486" , "smbr" , "9946567" };
String[] number_two = { "+122637867" , "486" , "nrkZone"  , "smbr"};


HashSet<String> matched = new HashSet<String>();
HashSet<String> mismatched = new HashSet<String>();

Now I have quite a few for loops, this part I figure you could simplify into maybe just one. But alas, it works. Now as to why . Well, the way this code works is by assuming all of the values in the arrays have no matches until proven otherwise. So it simply puts all of them into the mismatched HashSet.

for (int i = 0; i < number_one.length; i++) {
    mismatched.add(number_one[i]);
}

for (int i = 0; i < number_two.length; i++) {
    mismatched.add(number_two[i]);
}

Now you need to nest two for loops to iterate through both arrays, and in the process checking for any matches, and if there is one, it is added to the matched HashSet.

for(int n = 0; n < number_one.length; n++) {
       for (int m = 0; m < number_two.length; m++) {
           if(number_one[n].equals(number_two[m])){
            matched.add(number_one[n]);
           }
       }
   }

Now we simply remove all of the values that we know have matches from the mismatched variable, which already has ALL of the possible values. We are simply removing what we know matches.

for (int i = 0; i < matched.size(); i++) {
    mismatched.remove(matched.toArray()[i]);
}

And here we simply print them out in a semi-neat fashion.

System.out.print("Matched: ");
for (int i = 0; i < matched.size(); i++) {
    System.out.print(matched.toArray()[i] + " ");
}

System.out.println("");
System.out.print("Mismatched: ");
for (int i = 0; i < mismatched.size(); i++) {
    System.out.print(mismatched.toArray()[i] + " ");
}

Answer 2

Try this code implemented using array.

    String[] number_one = { "info@fn.ca" , "+122637867"  , "486" , "smbr" , "9946567" };
    String[] number_two = { "+122637867" , "486" , "nrkZone"  , "smbr"};
    boolean mismatch=true;
    for(int i=0;i<number_one.length;i++){
        mismatch=true;
        for(int j=0;j<number_two.length;j++){
        if(number_one[i].equals(number_two[j])){
            System.out.println(number_one[i]+"Matches");
        mismatch=false;
        break;
        }
        }
    if(mismatch){
        System.out.println(number_one[i]+"Mis Matches");
    }
    }