CODEWITHSUNDEEP
javaarraysmatrix
JacobIRR
JacobIRR

Reputation: 8946

Return integer index of 1D array for 2D array of coordinates pairs

After reading 10+ threads about converting an entire array from 2D to 1D and vice versa, I'm wondering if there is a mathematical formula, requiring no iteration to return the index position of a pair of integers for a sorted 2D array.

My "grid" is always square, can be any size (it's 3x3 in this example), and I've got a sorted block of human friendly values from which I need to retrieve what I'm calling "True 1D indices" like this:

"Human-friendly" coordinates| Java 2D Indices    |  True 1D indices
    [1,1],[1,2],[1,3],  ==>   [0,0],[0,1],[0,2], ==>  0 , 1 , 2
    [2,1],[2,2],[2,3],  ==>   [1,0],[1,1],[1,2], ==>  3 , 4 , 5
    [3,1],[3,2],[3,3],  ==>   [2,0],[2,1],[2,2], ==>  6 , 7 , 8

So I need a method for my class to give me the following results:

I enter 1,1 and get back 0, I enter 3,2 and get back 7, I enter 2,3 and get back 5, etc etc...

I've played around with half a dozen equations where I try things like adding the coordinates to the previous row index squared, and I can never get the right result for every cell in the grid. Is there some special operator or Math function I'm missing?

Thanks.

Upvotes: 3

Views: 1888

Answers (2)

bit-shashank
bit-shashank

Reputation: 921

let the matrix be of 3*3 ,3 rows and 3 column and we have to find the indices of (i,j) then the formula will be. indices= 3*(i-1)+j;

Note :- here 3*3 and i,j are not it java 2d array format it is in human friendly coordinates.

Example (i,j)=(2,3) indices=3*(2-1)+3 =6 And since your indices starts from 0 you could simply subtract one from 6 i.e. 5.

Upvotes: 2

dukr
dukr

Reputation: 26

If the indices are named [index1,index2],

1DIndex = (rowNumber*index1) + index2

Where rowNumber starts at 1 for the first row, and index1 and index2 are the Java indices.

Upvotes: 0

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