Appending Integer array elements in Java

Question

I have an array, say

int a[]={2,0,1,0,1,1,0,2,1,1,1,0,1,0,1};

I need to append each of the 5 neighboring elements and assign them to a new array b with length=(a.length/5); and i want to append the 5 neighboring elements so that I have: int b[]={20101,10211,10101}; I need to do this for various length arrays, in most cases with length of a being greater than 15.

Any help would be greatly appreciated, I'm programming in Java.

Thanks in advance.

Integer concatenation doesn't necessarily make sense -- for example, have you considered what behavior you'd like to define if the integers are negative? — Etaoin, Apr 28 '10 at 10:11
What do you want to happen when the first element of each 5-int-block is a 0? Should the result of 0,1,2,3,4 be 01234 or 1234? — Thomas Lötzer, Apr 28 '10 at 10:15
I need it to be 01234, since I'm using this for encoding messages, I need to keep all 5 digits. — Mohammad Sepahvand, Apr 28 '10 at 10:18
See also: http://stackoverflow.com/questions/2674707/how-to-concatenate-int-values-in-java/ — polygenelubricants, Apr 28 '10 at 10:40

score 4 · Accepted Answer · answered Apr 28 '10 at 10:11

4

It's pretty straighforward:

// Assuming a.length % 5 == 0.
int[] b = new int[a.length / 5];
for (int i = 0; i < a.length; i += 5) {
    b[i/5] = a[i]*10000 + a[i+1]*1000 + a[i+2]*100 + a[i+3]*10 + a[i+4];
}

answered Apr 28 '10 at 10:11

Marcelo Cantos

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1

If a.length does not divide evenly by 5 the array will not be big enough, and various exceptions will be thrown inside the for loop. – Finbarr Apr 28 '10 at 10:13
Thansk, that works fine, but when I'm dealing with integers starting with 0, with the above method I lose the 0, for example: a={0,2,2,2,2,1,0,2,0,2}; I get b={2222,10202}; instead of b={02222,10202}; which is what i need. Any way to stop that happening? – Mohammad Sepahvand Apr 28 '10 at 10:17
@Finbarr, the length of a will always be divisible by 5, so thankfully that won't be a problem. – Mohammad Sepahvand Apr 28 '10 at 10:19
4

If you want to keep the leading zeroes, you cannot store these as integers. Use Strings instead. – Syntactic Apr 28 '10 at 10:20
2

If you need to keep the leading zeroes, then b cannot be an `int[]`, because numbers can't have leading zeroes. You could make it for example a `String[]`. – Jesper Apr 28 '10 at 10:20

score 2 · Answer 2 · answered Apr 28 '10 at 10:14

This sounds like a homework question, so I won't give you the complete solution, but the basic rundown is:

Compute the length of b: len = a.length / 5
Construct b with that many elements.
Initialize an index variable to point to the first element in a
For each element in b:
- Construct the value for that element from a[idx]...a[idx+4]
- Advance the index into a by 5.

Also note that you may need to verify that the input a is actually a multiple of 5 in length.

polygenelubricants · Answer 3 · 2010-04-28T11:01:40.090

This works with (a.length % 5) != 0, and keeps leading zeroes (by storing digits into String).

    int a[]={2,0,1,0,1,1,0,2,1,1,1,0,1,0,1,0,0,7};

    final int N = 5;
    String b[] = new String[(a.length + N - 1)/ N];
    StringBuilder sb = new StringBuilder(N);
    int x = 0;
    for (int i = 0; i < b.length; i++) {
        sb.setLength(0);
        for (int k = 0; k < N && x < a.length; k++) {
            sb.append(a[x++]);
        }
        b[i] = sb.toString();
    }
    System.out.println(java.util.Arrays.toString(b));
    // prints "[20101, 10211, 10101, 007]"

Alternately, you can also use regex:

    String[] arr =
        java.util.Arrays.toString(a)
        .replaceAll("\\D", "")
        .split("(?<=\\G.{5})");
    System.out.println(java.util.Arrays.toString(arr));
    // prints "[20101, 10211, 10101, 007]"

Basically this uses Arrays.toString(int[]) to append all digits into one long String, then removes all non-digits \D, then uses \G-anchored lookbehind to split every .{5}

score 0 · Answer 4 · answered Oct 22 '12 at 14:09

import java.util.Arrays;

public class MainClass {
   public static void main(String args[]) throws Exception {
      int array[] = { 2, 5, -2, 6, -3, 8, 0, -7, -9, 4 };
      Arrays.sort(array);
      printArray("Sorted array", array);
      int index = Arrays.binarySearch(array, 1);
      System.out.println("Didn't find 1 @ "
      + index);
      int newIndex = -index - 1;
      array = insertElement(array, 1, newIndex);
      printArray("With 1 added", array);
   }
   private static void printArray(String message, int array[]) {
      System.out.println(message
      + ": [length: " + array.length + "]");
      for (int i = 0; i < array.length; i++) {
         if (i != 0){
            System.out.print(", ");
         }
         System.out.print(array[i]);         
      }
      System.out.println();
   }
   private static int[] insertElement(int original[],
   int element, int index) {
      int length = original.length;
      int destination[] = new int[length + 1];
      System.arraycopy(original, 0, destination, 0, index);
      destination[index] = element;
      System.arraycopy(original, index, destination, index
      + 1, length - index);
      return destination;
   }
}

It will print

Sorted array: [length: 10]
-9, -7, -3, -2, 0, 2, 4, 5, 6, 8
Didn't find 1 @ -6
With 1 added: [length: 11]
-9, -7, -3, -2, 0, 1, 2, 4, 5, 6, 8

score 0 · Answer 5 · answered Apr 28 '10 at 10:27

Naive approach.

import java.util.ArrayList;

/* Naive approach */
public class j2728476 {
    public static void main(String[] args) {
        int a[] = {2,0,1,0,1,1,0,2,1,1,1,0,1,0,1};
        ArrayList<String> al = new ArrayList<String>();
        String s = "";
        for (int i = 0; i < a.length; i++) {
            if (i % 5 == 0 && i != 0) {
                al.add(s);
                s = "" + a[i];
            } else {
                s += a[i];
            }
        }
        al.add(s);
        for (String t : al) {
            // convert values to ints ...
            System.out.println(t);
        }
    }
}

Will print:

20101
10211
10101

Appending Integer array elements in Java

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