Using an ArrayList to search within other ArrayList

Question

I cant seem to figure out a way to use an ArrayList (2 or more indexes) to search within another ArrayList. For example, if list = [4,5,6,2,3,5,6] and searchlist = [5,6], index should = 1.

Anybody have any ideas?

so are you looking for the occurrence of the numbers in the searchlist to be in your list in the order that it is in in searchlist? and it return the index? — mig, Mar 21 '14 at 04:39
you want to search the second list within the first list.. right..?? — Dileep, Mar 21 '14 at 04:44
See a similar post at http://stackoverflow.com/questions/3940194/find-an-array-inside-another-larger-array — Setyo N, Mar 21 '14 at 04:51
so what happens if 5, 6 are present but not together.. like 5,2,6 — Dileep, Mar 21 '14 at 04:53
Collections.binarySearch() method may be helpful--> http://docs.oracle.com/javase/7/docs/api/java/util/Collections.html#binarySearch(java.util.List, T) — MaheshVarma, Mar 21 '14 at 05:06

score 1 · Answer 1 · answered Mar 21 '14 at 04:48

Use pointers to your current indices

public int search(int[] list, int[] searchList) {
  int listIndex = 0;
  int searchListIndex = 0;
  int foundIndex = -1;

  while (listIndex != list.length) {
    if (list[listIndex] == searchList[searchListIndex]) {
      if (searchListIndex == 0) {
        foundIndex = listIndex;
      }
      listIndex++;
      searchListIndex++;
    }
    else {
      listIndex++;
      searchListIndex = 0;
      foundIndex = -1;
    }
  }
  return foundIndex;
}

score 1 · Answer 2 · answered Mar 21 '14 at 04:49

Loop through the arrays in a nested loop. Then compare all values. If all values match up, then return the index:

list = [4,5,6,2,3,5,6]
searchlist = [5,6]

limit = list.size - serachList.size - 1

#make sure list's size is greater than or equal to searchList's size
for(i = 0 to limit)
    for(j = 0 to searchList.size-1)
        if(list[i + j] != searchList[j]) #found mismatch, so move on
            break
        if(j = searchList.size-1) #found complete match
            return i

score 1 · Accepted Answer · answered Mar 21 '14 at 04:51

List has subList and equals. That's all you need.

import java.util.*;
import java.lang.*;
import java.io.*;


class TEST
{
public static void main (String[] args) throws java.lang.Exception  {
    List<Integer> list1 = Arrays.asList(1,2,3,4,5);
    List<Integer> list2 = Arrays.asList(2,3);

    for (int i=0; i<list1.size()-list2.size(); ++i) {
        if (list1.subList(i, i+list2.size()).equals(list2)) {
            System.out.println("found: "+i);
            return;
        }
    }

    System.out.println("not found");
}
}

score 0 · Answer 4 · answered Mar 21 '14 at 04:39

The simplest idea is to run a loop, get the first value of the searchlist and find that value in the other list. If you find the number, increment the indices of both the lists and check the values again, continue until the end of searchlist or until the end of the other list.

I am not posting the code as I would rather you try yourself first, and come back if you are stuck anywhere.

Using an ArrayList to search within other ArrayList

4 Answers4