How to insert an integer into a sorted List in Java?

First, you need to detect the sort order in the existing list.

1. If the list is empty, you don’t need to care, just add your element, you cannot break any existing sort order.
2. If the list contains one element, I cannot tell you what to do. Options include tossing a coin and throwing an exception, but there are more.
3. If the list contains two or more elements, iterate through them except for the last element, each time comparing the current element with the element in the next higher index. As soon as you encounter a pair of elements that are not equal, you know the sort order. If you only encounter equal elements, all the elements in the list are equal, and again I cannot tell you what to do.

Once you’ve detected a sort order, I suggest a big if-else statement to handle the two cases separately.

You already have the code for the ascending case. Except it doesn’t always work. If I try to insert 4 into { 1, 3, 5 }, I get an ArrayIndexOutOfBoundsException. If I try with { 1, 3, 5, 7, 9 }, the 9 is lost. You should probably find a fix for that.

You should be able to handle the descending case similarly to the ascending case. Only use num > list.get(i) instead of num < list.get(i).

First, you need to check whether the list is sorted in ASC or DESC. That's easy: compare the 1st two elements look for two consecutive non-identical elements and see whether the first one is greater or the second one (thanks tom for correcting the mistake).

Now, for a list sorted in ascending order, convert list to an ArrayList. Then, add num to the ArrayList and convert it to a TreeSet, since TreeSets sort their elements in ascending order. Finally, reassign the ArrayList to hold the contents of the TreeSet and return it.

For a list sorted in descending order, first, convert the TreeSet to an ArrayList, then iterate over this ArrayList in reverse order and add the elements to a new ArrayList. Return the second ArrayList.

public List<Integer> insertAndSort(List<Integer> list, int num){

    ArrayList<Integer> a = new ArrayList<Integer>(list);
    a.add(new Integer(num));
    TreeSet t = new TreeSet(a);
    a = new ArrayList<Integer>(t);

    int l = list.size();
    for(int i=1; i<l; i++){
        if(list.get(i) != list.get(i-1))
            break;
    }

    if(list.get(i) > list.get(i-1)) //ASC
        return a;
    else{ //DESC
        ArrayList<Integer> a2 = new ArrayList<Integer>();
        for(int i = a.size() - 1; i >= 0; i--)
            a2.add(a.get(i));
        return a2;
    }
}

I would proceed with something like the code below. A few remarks :

• it is possible to decide sort order if and only if there is at least 1 element AND first and last elements have different values.
• the line int oo = list.get(p2) - list.get(p1); compute the difference between the last and first element (negative = DESC, positive = ASC)
• the variable ox is positive if and only if the added element is after the element pointed by the variable p3 whatever the order.
• the position is found by a binary search algorithm by choosing p3 between p1 and p2 and deciding if the element is before or after p3.

This is not fully tested bu works with the examples you gave :

// static List<Integer> list = new ArrayList<>(Arrays.asList(7, 5, 3, 1));
static List<Integer> list = new ArrayList<>(Arrays.asList(1, 3, 5, 7));

static void add(int x)
{
    // fixed code below (see Ole V.V. comment)
}

static public void main(String [ ] args)
{
    add(4);

    for(int x: list)
        System.out.println(x);
}

EDIT: to be complete (see Ole V.V. comments)

• When the new element is to be inserted before the first or after the last, two simple O(1) tests may be performed ; the general case has O(log N) complexity, this is more efficient than traversing the entire list which is O(N).
• Be carefull too when comparing elements to deduce sort order, there may be several equal values ; the best is to compare the first and the last elements, this is O(1) - again better than O(N).

Algorithmic complexity is an important matter (to me) and was the primary aim of my post. The code of the add function becomes :

static void add(int x)
{
    int p1 = 0;
    int p2 = list.size() - 1;

    if(list.size() == 0 || list.get(p1) == list.get(p2))
        throw new IllegalStateException("Unable to decide sort order");

    int oo = list.get(p2) - list.get(p1);

    if(oo * (x - list.get(p1)) <= 0)       // new element is before first
        list.add(0, x);
    else if(oo * (x - list.get(p2)) >= 0)  // new element is after last
        list.add(x);
    else
    {
        while (p1 + 1 < p2)
        {
            int p3 = (p1 + p2) / 2;

            int ox = (x - list.get(p3)) * oo;

            if(ox >= 0)  // new element is after p3
                p1 = p3;

            if(ox <= 0)  // new element is before p3
                p2 = p3;
        }

        list.add(p2, x);
    }
}

Note : there may be still some undealed side effects. If the asker is interested, I am ready to give further help - just ask.

Your code and what you say is two different things. Based on code you made, I see that List can't contain more than one instance of same value. If thats the case, main method should look like this:

public static void main(String[] args){

  int num = 4;
    List<Integer> myList = Arrays.asList(1, 3, 4, 5, 7);
    List<Integer> newList;
    if (!myList.contains(num)) {
        newList = insertAndSort(myList, num);
    } else {
        newList = copyList(myList);
    }

    System.out.println(newList);

}

if thats not the case:

public static void main(String[] args){

  int num = 4;
    List<Integer> myList = Arrays.asList(1, 3, 4, 5, 7);
    List<Integer> newList;
    newList = insertAndSort(myList, num);

    System.out.println(newList);

}

Methods that main method use:

Assuming we can only know how list is sorted by values in list, we have to decide default sort method when there is only 1 value or all values are same. I picked ASC as default. If elements can be of same value and we are certain list is sorted its best to compare lowest value in list with the highest one. With this approach we have to compare 2 elements only once.

So method to see if list is sorted DESC would look like this:

private static boolean isDesc(List<Integer> list) {
    return list.get(0) > list.get(list.size() - 1);
}

If method returns true its sorted DESC. With returning false its ASC. If we want to change default value, when values don't tell us how its sorted and we want it to be DESC we would replace '>' sign with '>='

private static boolean isDesc(List<Integer> list) {
    return list.get(0) >= list.get(list.size() - 1);
}

Code for as you called it insertAndSort:

public static List<Integer> insertAndSort(List<Integer> list, int num) {
    List<Integer> listSecond = new ArrayList<Integer>(list.size() + 1);
    boolean isDescSortOrder = isDesc(list);
    if (isDescSortOrder) {
        int i = 0;
        while ((i < list.size()) && (list.get(i) > num)) {
            listSecond.add(list.get(i));
            i++;
        }
        listSecond.add(num);
        while (i < list.size()) {
            listSecond.add(list.get(i));
            i++;
        }
    } else { // is asc sort order
        int i = 0;
        while ((i < list.size()) && (list.get(i) < num)) {
            listSecond.add(list.get(i));
            i++;
        }
        listSecond.add(num);
        while (i < list.size()) {
            listSecond.add(list.get(i));
            i++;
        }

    }
    return listSecond;
}

Depending on how its sorted code got devided into 2 blocks. As we don't know on each iteration will be right place to insert our num value its better to use while loop. Using for loop and checking everytime if num value already exists in list is counterproductive. On top of that I dunno if your application should allow same values in the list. If I assume it should, you can't add num value if it already existed in the list with for loop and checking with contains every iteration.

And just to copy the table when list already has the element and we don't want to add elements that are already incuded:

public static List<Integer> copyList(List<Integer> list) {
        List<Integer> listSecond = new ArrayList<Integer>(list.size());
        for (int i = 0; i < list.size(); i++) {
            listSecond.add(list.get(i));
        }
        return listSecond;
    }

Also based on good programming practices I'd recommend to name methods based on what they do.

Calling method insertAndSort is asking for trouble. If I seen it, I'd say it alters list we are giving in parameter. Then it sorts it. So putting unsorted list in it would give us unwanted outcome. And I'd still ask myself why does it return a List when we are inserting an item to already existing list? I'd rather name it:

public static List<Integer> copySortedWithNewValue(List<Integer> sortedList, int newValue)

Try to use PriorityQueue. This collection contains a sorted sequence of elements which could be duplicated.