CODEWITHSUNDEEP
javasortingoutput
catch32
catch32

Reputation: 18572

Determines if the array list is sorted

I need to estimate if the array list is sorted (don't sort).

When Strings are sorted, they are in alphabetical order. I try to use compareTo() method to determine which string comes first

And return true if the array list is sorted, else false.

Code:

public boolean isSorted()
{
    boolean sorted = true;        
    for (int i = 1; i < list.size(); i++) {
        if (list.get(i-1).compareTo(list.get(i)) != 1) sorted = false;
    }

    return sorted;
}

Easy test:

    ArrayList<String> animals = new ArrayList<String>();
    ArrayListMethods zoo = new ArrayListMethods(animals); 
    animals.add("ape");
    animals.add("dog");
    animals.add("zebra");

    //test isSorted
    System.out.println(zoo.isSorted());
    System.out.println("Expected: true");

    animals.add("cat");
    System.out.println(zoo.isSorted());
    System.out.println("Expected: false");

    animals.remove("cat");
    animals.add(0,"cat");
    System.out.println(zoo.isSorted());
    System.out.println("Expected: false");

    **Output:**
    false
    Expected: true
    false
    Expected: false
    false
    Expected: false

This easy test shows only 1/3 coverage.

How to solve this issue.

Upvotes: 5

Views: 22576

Answers (6)

prasanth
prasanth

Reputation: 3602

You can write a utily method like isSortedList(List list).

public static boolean isSortedList(List<? extends Comparable> list)
{
    if(list == null || list.isEmpty())
        return false;

    if(list.size() == 1)  
        return true;

    for(int i=1; i<list.size();i++)
    {
        if(list.get(i).compareTo(list.get(i-1)) < 0 )
            return false;
    }

    return true;    
}

As as utility method, you can use it anywhere.

Upvotes: 1

Ruchira Gayan Ranaweera
Ruchira Gayan Ranaweera

Reputation: 35557

Try this

import java.util.ArrayList;
import java.util.Collections;
import java.util.List;

public class Sort {
public static void main(String []args) {
    List<String> l1=new ArrayList<String>();
    List<String> l2=new ArrayList<String>();
    l1.add("a");
    l1.add("b");
    l1.add("c");

    l2.add("b");
    l2.add("c");
    l2.add("a");

     if(isSorted(l1)){
         System.out.println("already sorted");
     }
    else{
         Collections.sort(l1);
     }
   }
public static boolean isSorted(List<String> list){
    String previous = "";
    for (String current: list) {
        if (current.compareTo(previous) < 0)
            return false;
        previous = current;
    }
    return true;
}
}

Upvotes: 1

Saher Ahwal
Saher Ahwal

Reputation: 9237

You need to check for unsorted case.

This means that if you are assuming sorting ascending, the unsorted case will be finding an element at index i being out of order from index i-1 where element[i] < element[i-1].

Upvotes: 0

Isaac.Lopez
Isaac.Lopez

Reputation: 65

you have to change the compareTo expresion to any positive number, which indicates that the previous element is alphabetically after the current element, hence the list is not ordered 

    public boolean isSorted()
    {
        boolean sorted = true;        
        for (int i = 1; i < list.size(); i++) {
            if (list.get(i-1).compareTo(list.get(i)) > 0) sorted = false;
        }

        return sorted;
    }

Upvotes: 0

AllTooSir
AllTooSir

Reputation: 49372

Change the condition :

if (list.get(i - 1).compareTo(list.get(i)) >0)

You should check for >0 instead of !=-1 .

Go through the documentation of compareTo()

the value 0 if the argument string is equal to this string; a value less than 0 if this string is lexicographically less than the string argument; and a value greater than 0 if this string is lexicographically greater than the string argument.

Upvotes: 2

Grisha Weintraub
Grisha Weintraub

Reputation: 7986

You have a small bug in your method. Should be :

public boolean isSorted()
{
    boolean sorted = true;        
    for (int i = 1; i < list.size(); i++) {
        if (list.get(i-1).compareTo(list.get(i)) > 0) sorted = false;
    }

    return sorted;
}

>0 instead of !=1, you can't be sure that 1 is returned..

Upvotes: 8

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