Array search code

Question

In the book 'Data structures and algorithms in java' the following Array search method code is provided:

{
    int j;
    for(j=0; j< nElems; j++)            // for each element,
        if( a[j].getLast().equals(searchName))  // found item?
             break;                       // exit loop before end
    if(j == nElems)                    // gone to end?
      return null;                    // yes, can't find it
    else
      return a[j];                    // no, found it
}

I am trying to understand why there needs to be a if(j == nElems) check? Wouldn't the method work the same if it were written as:

{           
    int j;
    for(j=0; j <nElems; j++)               
      if( a[j].getLast().equals(searchName))      
         return a[j];              
    return null;
}

You could understand the code better if format it correctly. — Moha Dehghan
– Moha Dehghan, Commented Feb 19, 2013 at 12:06

Caesar Ralf · Accepted Answer · 2013-02-19 12:00:37Z

1

it would :P You could declare the j inside the for to limit its scope as well.

In the first implementation what it do is check if it iterated over all elements and didn't find anything, because j was incremented until it is equal the the condition of stop of the for-loop. I.e., it didn't stop because of break, indicating that it found an element.

I prefer your solution because it's easier to read :)

answered Feb 19, 2013 at 12:00

Caesar Ralf

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Comments

Arun P Johny · Accepted Answer · 2013-02-19 11:59:58Z

1

Yes, both of these ways are give the same result.

answered Feb 19, 2013 at 11:59

Arun P Johny

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Comments

Caffe Latte · Accepted Answer · 2013-06-07 08:06:42Z

0

Well , j never equals to nElems so this condition (j==nElems) is not wrong BUT it's not Working . You can make it like this (j==nElems-1) but it's steal waste of code , So your algorithm is better than the first one .

edited Jun 7, 2013 at 8:06

Caffe Latte

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answered Feb 19, 2013 at 12:29

Azad

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