Well simple question here (maybe not a simple answer?)
Say I have a two dimensional array
[0] [1] [2]
[3] [4] [5]
[6] [7] [8]
Now suppose I want to get the position of the number 6
I know with a one-dimensional array i can use Array.indexOf() but what would my options be with 2-dimensional arrays?
Thanks!
Linear Search in 2D Array: It is used to find whether a particular element is present in the array or not by traversing every element in the array. While searching in the 2D array is exactly the same but here all the cells need to be traversed In this way, any element is searched in a 2D array.
The elements of a 2D array are arranged in rows and columns, and the new operator for 2D arrays specifies both the number of rows and the number of columns. For example, int[][] A; A = new int[3][4]; This creates a 2D array of int that has 12 elements arranged in 3 rows and 4 columns.
Accessing 2D Array Elements In Java, when accessing the element from a 2D array using arr[first][second] , the first index can be thought of as the desired row, and the second index is used for the desired column. Just like 1D arrays, 2D arrays are indexed starting at 0 .
I'd say something like this:
public static Tuple<int, int> CoordinatesOf<T>(this T[,] matrix, T value)
{
int w = matrix.GetLength(0); // width
int h = matrix.GetLength(1); // height
for (int x = 0; x < w; ++x)
{
for (int y = 0; y < h; ++y)
{
if (matrix[x, y].Equals(value))
return Tuple.Create(x, y);
}
}
return Tuple.Create(-1, -1);
}
Here is a method that should find an index in an array with an arbitrary rank.
... Added Upper/Lower bounds range per rank
public static class Tools
{
public static int[] FindIndex(this Array haystack, object needle)
{
if (haystack.Rank == 1)
return new[] { Array.IndexOf(haystack, needle) };
var found = haystack.OfType<object>()
.Select((v, i) => new { v, i })
.FirstOrDefault(s => s.v.Equals(needle));
if (found == null)
throw new Exception("needle not found in set");
var indexes = new int[haystack.Rank];
var last = found.i;
var lastLength = Enumerable.Range(0, haystack.Rank)
.Aggregate(1,
(a, v) => a * haystack.GetLength(v));
for (var rank =0; rank < haystack.Rank; rank++)
{
lastLength = lastLength / haystack.GetLength(rank);
var value = last / lastLength;
last -= value * lastLength;
var index = value + haystack.GetLowerBound(rank);
if (index > haystack.GetUpperBound(rank))
throw new IndexOutOfRangeException();
indexes[rank] = index;
}
return indexes;
}
}
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With