Check if array is sorted in java

Question

I want to create a method that returns 0 if array not sorted returns 1 if it is sorted in ascending order and returns -1 if its sorted in descending order

This is what I have done so far:

public static int isSorted(int[] intArray) {

    int end = intArray.length - 1;
    int val = 0;

    for (int i = 1; i < end; i++) {

        if (intArray[0] < intArray[i]) {

            val = 1;
        }

        else if (intArray[0] > intArray[i]) {

            val = -1;
        }
    }

    return v;
}

}

This returns 1 if its ascending and -1 if its descending. But if I create a random array it does not return 0. The question is how to check if both conditions fail, i.e., if its not sorted at all.

go step by step into your code for array 1 2 1 and write down value of val variable at every loop iteration — Iłya Bursov
– Iłya Bursov, Commented Apr 17, 2017 at 19:59
Just a note: In your for-loop replace i < end with i <= end since end is the length but already subtracted one — Schokokuchen Bäcker
– Schokokuchen Bäcker, Commented Apr 17, 2017 at 20:01
If the first two elements in an array are increasing, check that all subsequent elements are increasing too; if the first two elements are decreasing, check that all subsequent elements are decreasing too. — Andy Turner
– Andy Turner, Commented Apr 17, 2017 at 20:01
When you say "ascending" and "descending" order, do you actually mean that, or do you mean "non-descending" and "non-ascending"? (e.g. 1,2,3 is ascending, 1,1,2 is non-descending). — Andy Turner
– Andy Turner, Commented Apr 17, 2017 at 20:37
it takes just as much code to sort an array as it does to check if its sorted....so why not just sort it by default...am i wrong? — Tim Boland
– Tim Boland, Commented Jul 12, 2017 at 1:23

K.Rzepecka · Accepted Answer · 2017-04-17 20:31:04Z

2

You can use two additional variables to count:

public static int isSorted(int[] intArray) {

    int end = intArray.length-1;
    int counterAsc = 0;
    int counterDesc = 0;

    for (int i = 0; i < end; i++) {
        if(intArray[i] < intArray[i+1]){
            counterAsc++;
        }
        else if(intArray[i] > intArray[i+1]){
            counterDesc++;
        }
    }
    if(counterDesc == 0){
        return 1;
    }
    else if(counterAsc == 0){
        return -1;
    }
    else return 0;
}

edited Apr 17, 2017 at 20:31

answered Apr 17, 2017 at 20:17

K.Rzepecka

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Comments

shri cool · Accepted Answer · 2017-04-17 20:14:26Z

1

Your comparison is just between first element and other elements in the array.

val = 0;
for(int i=0;i<=end;i++){
     for(int j=0;j<end;j++){
          if(intArray[j]>intArray[j+1])
               val = 1; 

     }
   }
 return val;

answered Apr 17, 2017 at 20:14

shri cool

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Comments

Ammar Akouri · Accepted Answer · 2017-04-17 20:25:57Z

Here you go

public class stackoverflow {


    public static int isSorted(int[] intArray) {


    boolean sortedAsc = true;
    boolean sortedDesc = true;
    boolean sameValues=true;
    int result = 0;

    for (int i = 0; i < intArray.length-1; i++)
    {

        if (intArray[i] > intArray[i+1]) {
            sortedAsc=false;
            sameValues=false;
        }
        if (intArray[i] < intArray[i+1]) {
            sortedDesc=false;
            sameValues=false;
        }
        }

    if(sortedAsc) result= 1;
    if(sortedDesc) result =  -1;
    if(sameValues) result = 2;

    return result;
}

    public static void main(String[] args) {
        // TODO code application logic here
        int array[] = new int[4];
        array[0]=1;
        array[1]=2;
        array[2]=3;
        array[3]=4;

        System.out.println(isSorted(array));

    }

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