In C#, Rotate a List to The Right by the specified numbers of places without using LINQ?

Question

I am trying to rotate a List of items to the right by the specified numbers of places without using LINQ and doing it the manual way, how am I able to do this?

I think I am not understanding how to approach/solve this problem in particular. Here's what I have tried so far.

A clear example of this problem could be something like this: Initial Array(or List): 20,30,40,50,60,70 Rotate to the right by 3 places: 50,60,70,20,30,40

      public void Test(List<int> items, int places)
    {
        items.RemoveAt(places);
        items.Add(places);

    }

It will be more clear if you include an example, of list before this operation and list after this operation — sujith karivelil, Jul 19 '16 at 04:38
You need circular queue data structure. Add items in circular queue and just keep moving the start position as required. — Sarvesh Mishra, Jul 19 '16 at 04:44
A clear example of this problem could be something like this: Initial Array(or List): 20,30,40,50,60,70 Rotate to the right by 3 places: 50,60,70,20,30,40 — Zerox, Jul 19 '16 at 04:44
@Zerox the example you gave is a very special vase as it gives the same result for shifting 3 to the right and 3 to the left. I think it would have been better if you gave an example shifting 2 or 4 to the right. — Zein Makki, Jul 19 '16 at 05:33

Zein Makki · Answer 1 · 2016-07-19T05:51:15.903

Results from the below:

20,30,40,50,60,70 
60,70,20,30,40,50  (Rotate 2) 
50,60,70,20,30,40  (Rotate 3)

If you meant to shift each element n indexes to the left, just replace the line list[i] = copy[index]; with list[index] = copy[i];

Then you get these results:

20,30,40,50,60,70 
40,50,60,70,20,30  (Rotate 2) 
50,60,70,20,30,40  (Rotate 3)

This is a simple working generic method:

static void RotateList<T>(IList<T> list, int places)
{
    // circular.. Do Nothing
    if (places % list.Count == 0)
        return;

    T[] copy = new T[list.Count];
    list.CopyTo(copy, 0);

    for (int i = 0; i < list.Count; i++)
    {
        // % used to handle circular indexes and places > count case
        int index = (i + places) % list.Count;

        list[i] = copy[index];
    }
}

Usage:

List<int> list = new List<int>() { 20, 30, 40, 50, 60, 70 };
RotateList(list, 3);

Or since I'm a big fan of extension Methods, you can create one:

(By using IList, this method works for arrays also)

public static class MyExtensions
{
    public static void RotateList<T>(this IList<T> list, int places)
    {
        // circular.. Do Nothing
        if (places % list.Count == 0)
            return;

        T[] copy = new T[list.Count];
        list.CopyTo(copy, 0);

        for (int i = 0; i < list.Count; i++)
        {
            int index = (i + places) % list.Count;

            list[i] = copy[index];
        }
    }

Usage:

List<int> list = new List<int>() { 20, 30, 40, 50, 60, 70 };
list.RotateList(12); // circular - no changes

int[] arr = new int[] { 20, 30, 40, 50, 60, 70 };

arr.RotateList(3);

should mod `places` by list length initially, else a `places` many times larger than list length will malfunction — zeromus, Jul 19 '16 at 04:54

score 4 · Answer 2 · answered Jul 19 '16 at 04:46

4

assuming you have a list 1-2-3-4-5, and you want to shift it 2 right to : 3-4-5-1-2.

for n = 1 to 2 
   remove the head, put it at the tail

simple loop with no linq.

Even though this sounds more like homework to me to be honest ..

answered Jul 19 '16 at 04:46

Noctis

11,507
3
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as an afterthought, you can use an array, create a new one with same size, and simply copy with an offset ... – Noctis Jul 19 '16 at 04:48

score 2 · Answer 3 · answered Jul 19 '16 at 05:16

2

int[] numbers = {20, 30, 40, 50, 60, 70};
int rotateBy = 3;
int[] rotated = new int[numbers.Length];

Array.Copy(numbers, rotateBy, rotated, 0, numbers.Length-rotateBy);
Array.Copy(numbers, 0, rotated, numbers.Length - rotateBy, rotateBy);

answered Jul 19 '16 at 05:16

Cetin Basoz

22,495
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score 0 · Answer 4 · edited May 23 '17 at 10:27

0

While this may not exactly answer your question it's worth mentioning about queues. For circular operations Queue<T>s could be a better data structure choice. To shift places it could be as simple as:

public static void Rotate<T>(this Queue<T> items, int places)
{
    for (int i = 0; i < places; i++)
        items.Enqueue(items.Dequeue());
}

E.g.

var q = new Queue<int>(new[] { 20, 30, 40, 50, 60, 70 });
q.Rotate(3);

Queues are also a lot more efficient than List<T>s for dequeuing operation since it doesnt have to remove from top and push entire block of array one up (read about it here: Queue<T> vs List<T>) or arrays which involves copying of the entire array.

edited May 23 '17 at 10:27

Community

1
1

answered Jul 19 '16 at 05:33

nawfal

70,104
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Isn't this shifting to the left ? Try `q.Shift(2);` – Zein Makki Jul 19 '16 at 05:35
@user3185569 I get `40,50,60,70,20,30`. Isn't that what's expected? – nawfal Jul 19 '16 at 05:38
shouldn't that be `60,70,20,30,40,50` ? Shifting each element 2 places **To the Right** ? Or am I missing something here ? – Zein Makki Jul 19 '16 at 05:39
@user3185569 Ah who knows. In that case it's a matter of adjusting the for loop. – nawfal Jul 19 '16 at 05:45

In C#, Rotate a List to The Right by the specified numbers of places without using LINQ?

4 Answers4