Java for loop - code efficiency

Question

I have an array of 10,000 elements and i want to loop through and find a particular number say '6573'.

Example

 for(int i=0;i<=user.length;i++)
    {
        if(user[i]=="6573")
            System.out.println("Found it!!!");
    }

Could you please suggest a way to improve the performance of the code.

Thanks

This won't work, use `i < user.length` instead. Remember index starts at `0`! — Christian Tapia, Jan 23 '14 at 05:31
this question belongs to [Code Review](http://codereview.stackexchange.com/) — Baby, Jan 23 '14 at 05:31
You can never find that element with the current way. Always use `equals()` method for String value comparisons, the condition in the `for` loop must be `i < user.length`, if you want to avoid `AIOOBE` and lastly, try to use a Map if you want faster retrievals(though inserts would be time consuming). — Rahul, Jan 23 '14 at 05:33

score 2 · Answer 1 · answered Jan 23 '14 at 05:33

Use .equals instead of == to compare strings
Break when a match is found
he end condition must be i < user.length to prevent ArrayIndexOutOfBoundsException:

--

for(int i = 0; i < user.length; i++) {
    if("6573".equals(user[i])) {
        System.out.println("Found it!!!");
        break;
    }
}

Note that I inverted the .equals() call to prevent NullPointerException if the array contains null values.

score 2 · Answer 2 · answered Jan 23 '14 at 05:46

If you need to do it once then that's about it. If you try to find several users in that list then you could use a set for O(1) search:

Set<String> set = new HashSet<>(Arrays.asList(user));
if(set.contains("6573"))
        System.out.println("Found it!!!");

And it may actually make sense to store your users directly in that set in the first place instead of using an array.

score 0 · Answer 3 · answered Jan 23 '14 at 05:35

0

if the array elements is in order(sorted) then u can use the binary search .. it will increase the performance of your program.

answered Jan 23 '14 at 05:35

Abhi Urs

95
3
8

score 0 · Answer 4 · answered Jan 23 '14 at 06:04

0

Sort the Array using sort() (efficiency O(n Log n) and use binary search (O(log n) ) .. I think this will be more efficient than your current efficency i.e, O(n).. Just giving details of @abhi's answer...

answered Jan 23 '14 at 06:04

TheLostMind

35,966
12
68
104

score 0 · Answer 5 · edited Jan 23 '14 at 06:51

0

try like this

int index = -1;  
boolean found = false;  
for(int i = 0; i < array.length; i++)  
{  
   if(array[i].equalsIgnoreCase(userInput))  
   {  
       index = i;
       found = true;
       break;
   }  
}

edited Jan 23 '14 at 06:51

dodgy_coder

12,407
10
54
67

answered Jan 23 '14 at 06:26

Shyam

6,376
1
24
38

score 0 · Answer 6 · answered Jul 05 '14 at 14:05

0

you could use multi-threading and make for example 2 pointers, one travel from the start of the array and one form the end till the middle of the array.
It will use more processing but less time.

answered Jul 05 '14 at 14:05

Baron Leonardo

329
3
13

Java for loop - code efficiency

6 Answers6