Java Finding a value in arraylist

Question

I have added few number into the arraylist . I would want to find the certain value from it.Example i have 4,4,9,9,18. I would like to find the value of 26. If 26 > largest value in the list it will display 18 and if value is 17 it will display 9, and if value is 5 it will display 4. Also is there another method to implement this search because liner search might be slow.

search value 26

    [4,4,9,9,18] display 18
    [20,20,29,29,4] display 20
    [28,28,28,1,10] display 28

if you have this list and search 26, it will output the first element. because the first element is <= than the value being search.

but current output is

Value of value2 : 9

    public class Arraylist {

    public static ArrayList<Integer> aList;

    public static void main(String[] args) {
        aList = new ArrayList<Integer>();
        aList.add(4);
        aList.add(4);
        aList.add(9);
        aList.add(9);
        aList.add(18);
        int value = 26;
        int value2 = 0;

        for (int i = 0; i < aList.size(); i++) {
            if (aList.get(i) <= value) {          
                if (i + 1 < aList.size()) {
                    value2 = aList.get(i);
                } else if(i > aList.size()) {
                    value2 = aList.get(i);

                }
            }
        }
        System.out.println("Value of value2 : " + value2);
    }
}

I do not understand your logic. Are you looking for a place where a given value should be inserted into a sorted ArrayList to keep it sorted? — PM 77-1
– PM 77-1, Commented Oct 25, 2013 at 4:05
Another method could be to start search from the middle of the list, and check if the variable is higher or lower than value in the array, if lower then search backwards from this point and if higher search forwards. — Scott Pearson
– Scott Pearson, Commented Oct 25, 2013 at 4:05
So you've been asked to implement something that sounds almost like binary search... — vanza
– vanza, Commented Oct 25, 2013 at 4:08
If you array is always sorted then binary search will be faster. — PM 77-1
– PM 77-1, Commented Oct 25, 2013 at 4:08

Sage · Accepted Answer · 2013-10-25 05:24:49Z

1

I have written the code using an array. You can easily adopt it to ArrayList

int a[] = {28,28,28,1,10};
// int a[] = {20,20,29,29,4}; // other input of yours
// int a[] = {4,4,9,9,18};  

   int x = 26;

   int liVal = -1;
   for(int i=0; i<a.length;i++)
       if(x < a[i]) // if we met a value > x
       {
          if(liVal==-1) // if we could not find any largest value smaller than x
              liVal = a[i]; // return the value > x
          break;
       }
       else if(x > a[i]) // find the largest value smaller than x, 
       {
           if(liVal < a[i])
               liVal = a[i];
       }

System.out.println(liVal);

edited Oct 25, 2013 at 5:24

answered Oct 25, 2013 at 5:17

Sage

15.5k3 gold badges36 silver badges40 bronze badges

Sign up to request clarification or add additional context in comments.

Comments

Aurand · Accepted Answer · 2013-10-25 04:05:50Z

0

A trivial and un-optimized version:

int value = 26 // or whatever parameter you get
int retVal = Integer.MIN_VALUE;
for (int i : list) {
  if (i <= value && i > retVal) {
    retVal = i;
  }
}
return retVal;

answered Oct 25, 2013 at 4:05

Aurand

5,5571 gold badge28 silver badges35 bronze badges

1 Comment

bcorso Over a year ago

This won't handle the case where the value < min(list) because the if statement is never entered since i <= value is never true. In that case you will return Integer.MIN_VALUE instead of min(list).

Itay Karo · Accepted Answer · 2013-10-25 04:06:34Z

0

If the array is sorted you can use a variant of binary search.
see http://en.wikipedia.org/wiki/Binary_search_algorithm

answered Oct 25, 2013 at 4:06

Itay Karo

18.4k4 gold badges43 silver badges59 bronze badges

Comments

Vidya · Accepted Answer · 2013-10-25 04:08:29Z

0

Once you sort the list, binarySearch in Collections will do the trick:

Collections.sort(aList)
int index = Collections.binarySearch(aList)

If index is nonnegative, the number was found in the list, and index is the position. If it is negative, it wasn't found, but index indicates where it would be if it were in the list.

And with O(log n) runtime for the search.

answered Oct 25, 2013 at 4:08

Vidya

30.4k7 gold badges48 silver badges73 bronze badges

2 Comments

Aurand Over a year ago

If you're only going to search a list once, the actual runtime is O(n log n) since you have to sort first.

Vidya Over a year ago

I was just referring to the search since that is what the OP seemed concerned about. Edited for clarity.

Francis · Accepted Answer · 2013-10-25 04:48:45Z

0

If I understand correctly, you want to find the biggest number in your array that is smaller or equal to value. I would do it like this:

for (int i = 0; i < aList.size(); i++) {
    if ( aList.get(i) <= value && aList.get(i) > value2) {
        value2 = aList.get(i);
    }
}

Also in your example, you do value2 = 0. This is ok if you can guarante that the array only contains positive values. Otherwise it would be better to use value2 = Integer.MIN_VALUE.

Finally, this code assume that the array is not guarantied to be sorted and that you will only need to search it once. Otherwise, a binary search could be more performant. Other answers on this question already show how to accomplish that.

edited Oct 25, 2013 at 4:48

answered Oct 25, 2013 at 4:06

Francis

1,0907 silver badges12 bronze badges

5 Comments

PM 77-1 Over a year ago

@user2822351 : If your ArrayList is unsorted, then straight forward search could be your only option.

PM 77-1 Over a year ago

How is value2 initialized? What will happen if value is bigger than all elements of the list?

BeyondProgrammer Over a year ago

the search suppose to search from [0]-[4] meaning from first number to the last, if value is bigger than all element, it will be the largest being displayed, if the value is =< than first element it will be the largest element in the list example [28,28,28,1,10] and you search 26

Francis Over a year ago

Edited for clarity. Indeed here I assumed that the list was not necessarily sorted. I also did not originally commented on the initial value of the value2 variable since the OP was concerned with the search.

BeyondProgrammer Over a year ago

I am more concerned of getting the correct display value and than the efficiency of the search

bcorso · Accepted Answer · 2013-10-25 05:22:03Z

0

According to the OP's comments:

The list isn't sorted
if value < min return min
if value > max return max

if min <= value <= max return nearest val <= value

public static int findValue(List<Integer> list, int value){
    int min = Integer.MAX_VALUE, nearest = Integer.MIN_VALUE;
    for(Integer v : list){
        if(v == value)
            return value;
        if(v > nearest && v < value)
            nearest = v;
        if(v < min)
            min = v;
    }
    return value < min ? min : nearest;
}

As a side note, you don't need to keep track of the max because nearest = max if value > max(list).

edited Oct 25, 2013 at 5:22

answered Oct 25, 2013 at 4:36

bcorso

47.5k10 gold badges66 silver badges76 bronze badges

2 Comments

BeyondProgrammer Over a year ago

I have added 3 example of list

bcorso Over a year ago

In your examples the exact rules you are trying to convey are ambiguous. I think you meant the last example to display the number immediately < 26 which is 10. In that case this code will give you the correct answer.

Collectives™ on Stack Overflow

Java Finding a value in arraylist

6 Answers 6

Comments

1 Comment

Comments

2 Comments

5 Comments

2 Comments

Your Answer

Linked

Hot Network Questions

Collectives™ on Stack Overflow

6 Answers 6

Comments

1 Comment

Comments

2 Comments

5 Comments

2 Comments

Your Answer

Sign up or log in

Post as a guest

Linked

Related