How to rotate a matrix in an array in javascript

Question

(disclosure, I\'m mostly math illiterate).

I have an array in this format:

var grid = [
  [0,0], [0,1], [0,2], [0,3],
  [1,0], [1,1], [1,2], [1,3],
          


        

Answer 1

These are two function for clockwise and counterclockwise 90-degree rotation:

    function rotateCounterClockwise(a){
        var n=a.length;
        for (var i=0; i<n/2; i++) {
            for (var j=i; j<n-i-1; j++) {
                var tmp=a[i][j];
                a[i][j]=a[j][n-i-1];
                a[j][n-i-1]=a[n-i-1][n-j-1];
                a[n-i-1][n-j-1]=a[n-j-1][i];
                a[n-j-1][i]=tmp;
            }
        }
        return a;
    }

    function rotateClockwise(a) {
        var n=a.length;
        for (var i=0; i<n/2; i++) {
            for (var j=i; j<n-i-1; j++) {
                var tmp=a[i][j];
                a[i][j]=a[n-j-1][i];
                a[n-j-1][i]=a[n-i-1][n-j-1];
                a[n-i-1][n-j-1]=a[j][n-i-1];
                a[j][n-i-1]=tmp;
            }
        }
        return a;
    }

Answer 2

Those looking for Rotating a two dimentional matrix (a more general case) here is how to do it.

example: Original Matrix:

[
  [1,2,3],
  [4,5,6], 
  [7,8,9]
]

Rotated at 90 degrees:

[
    [7,4,1]
    [8,5,2]
    [9,6,3]
]

This is done in following way:

matrix[0].map((val, index) => matrix.map(row => row[index]).reverse())

Answer 3

Credit goes to this answer for the actual rotation method.

My method was pretty straightforward. Just determine what the row length was, and then iterate through each item, converting the array index to x/y equivalents and then apply the method used in the linked answer to rotate. Finally I converted the rotated X/Y coordinates back to an array index.

var grid = [
  [0,0], [0,1], [0,2], [0,3],
  [1,0], [1,1], [1,2], [1,3],
  [2,0], [2,1], [2,2], [2,3],
  [3,0], [3,1], [3,2], [3,3]
]; 

var newGrid = [];
var rowLength = Math.sqrt(grid.length);
newGrid.length = grid.length

for (var i = 0; i < grid.length; i++)
{
    //convert to x/y
    var x = i % rowLength;
    var y = Math.floor(i / rowLength);

    //find new x/y
    var newX = rowLength - y - 1;
    var newY = x;

    //convert back to index
    var newPosition = newY * rowLength + newX;
    newGrid[newPosition] = grid[i];
}

for (var i = 0; i < newGrid.length; i++)
{   
    console.log(newGrid[i])
}

The output:

[3, 0] [2, 0] [1, 0] [0, 0]  
[3, 1] [2, 1] [1, 1] [0, 1]  
[3, 2] [2, 2] [1, 2] [0, 2]  
[3, 3] [2, 3] [1, 3] [0, 3]  

Fiddle for the lazy. And a 5x5 grid fiddle to demonstrate that the algorithm works for N grid sizes as long as they are square.

Answer 4

I don't really need to deal with indices, since I can copy the values from one place to the other, this simplifies the answer a bit:

var grid = [
  [0,0], [0,1], [0,2], [0,3], [0,4],
  [1,0], [1,1], [1,2], [1,3], [1,4],
  [2,0], [2,1], [2,2], [2,3], [2,4],
  [3,0], [3,1], [3,2], [3,3], [3,4],
  [4,0], [4,1], [4,2], [4,3], [4,4]
]; 

var side = Math.sqrt(grid.length);

var rotate = function(d,i){
   return [Math.abs(i % side - side+1), Math.floor(i/side)]
}
grid = grid.map(rotate);

You can see a jsfiddle here: http://jsfiddle.net/KmtPg/