I have a String array of size 1000:
String[] subStrStore = new String[1000];
I have only 6 elements in the array. If I sort it using:
Arrays.sort(subStrStore);
I am getting a null pointer exception because it tries to compare null elements. How can solve this problem?
If the strings are the first 6 elements you can use
Arrays.sort(subStrStore, 0, 6);
Use custom comparator,
Arrays.sort(subStrStore, new Comparator<String>() {
@Override
public int compare(String o1, String o2) {
if (o1 == o2)
return 0;
if (o1 == null) {
return -1;
}
if (o2 == null) {
return 1;
}
return o1.compareTo(o2);
}
});
this should push all your nulls at the end of your array
Remove the null elements and sort.
String[] subStrStore = new String[1000];
subStrStore[10] = "f";
subStrStore[200] = "e";
subStrStore[300] = "d";
subStrStore[500] = "c";
subStrStore[750] = "b";
subStrStore[900] = "a";
String[] sorted = Stream.of(subStrStore)
.filter(s -> s != null)
.sorted()
.toArray(String[]::new);
System.out.println(Arrays.toString(sorted));
// -> [a, b, c, d, e, f]
Have a look at Arrays.copyOf() from java.util.Arrays. It will do the job I believe .
I think the best way is to check element before compare. something like that:
if(object == null){break;}
THIS WILL WORK ONLY IF U USE FIRST ELEMENTS OF ARRAY. 1, 2, 3, 4 NULL, NULL ... x1000. NOT 1,2, NULL, NULL, 3, 4, 5. <- IF THAT WAY U HAVE TO USE EXCEPTIONS :) you dont want check other elements after first null, or you can use just try catch and handle the exception.
