I\'m attempting to create a method that checks an array for increasing elements. True should be returned if all the elements are in increasing order. I get an out-of-bounds
You get that exception as when (i+1)'s value becomes array.length. For example if you have an array of length 10, the elements indexes will be from 0,1,2...till 9. so either you have to check till i < arr.length - 1
or you can modify your logic accordingly.
I suggest you write your method like this
public static boolean isIncreasing(int[]arr)
{
for(int i=1; i<arr.length;i++)
{
if(arr[i-1]>arr[i])
return false;
}
return true;
}
it will help
Because in a list with n
items there are only n-1
gaps between them.
Change to
for (int i=0; i<arr.length-1; i++)
(Also you might want to check whether starting with false
and setting to true
is the right way around).
You can use Java 8's IntStream.
import java.util.stream.IntStream;
public class Test {
public static boolean isIncreasing(int[] a) {
return IntStream.range(1, a.length).reduce(0, (acc, e) -> acc + (a[e - 1] <= a[e] ? 0 : 1)) == 0;
}
}
You have two problems:
i+1
, i needs to finished incrementing one iteration earlier than a usual loop.{1, 2, 0}
when tested the first iteration tests 1 < 2
which is true, so return true - this is not what we want)Fixing these two problems:
int[] one = {1,2,3,4,5};
public static boolean isIncreasing(int[] arr) {
for(int i=0 ; i < arr.length - 1; i++) { // finish at length - 1
if (arr[i] > arr[i+1]) {
return false; // found elements that are out of order - return false
}
}
return true; // nothing out of order found - return true
}
This kind of logic - with an early exit with false on problem and a final return of true - is very common and a good pattern to learn.