CODEWITHSUNDEEP
javamatrix
user4833678
user4833678

Reputation:

Matrix rotation

folks, I've been analyzing the following code about the rotation of matrix by 90 degrees: 

static int[][] rotateCW(int[][] mat) {
    final int M = mat.length;
    final int N = mat[0].length;
    int[][] ret = new int[N][M];
    for (int r = 0; r < mat.length; r++) {
        for (int c = 0; c < mat[0].length; c++) {
            ret[c][mat.length-1-r] = mat[r][c];
        }
    }
    return ret;
}

The code works fine. What I do not understand is why c < mat[0].length Shouldn't it be c < mat[i].length ?

Upvotes: 0

Views: 526

Answers (4)

mazhar islam
mazhar islam

Reputation: 5619

The code should be:

for (int r = 0; r < M; r++) {
    for (int c = 0; c < N; c++) {
        ret[c][M-1-r] = mat[r][c]; 
        ...

Because using M and N is faster, saving indirection of storage access.

Also note that:

final int M = mat.length; // number of rows
final int N = mat[0].length; // number of columns

Because the number of columns in any row of any matrix is equal.

In the for loops ret[c][mat.length-1-r] or ret[c][M-1-r] is the last-to-first column for each row, where mat[r][c] is the is the first-to-last column for each row

Upvotes: 1

Sanjay Rabari
Sanjay Rabari

Reputation: 2081

mat[0].length representing the column length.

Upvotes: 0

Tom Jonckheere
Tom Jonckheere

Reputation: 1648

They are checking the length of the matrix. Since the length is the same, the solution is correct. It certainly shouldn't be c < mat[i].length because i is not known in the scope, it should be (as it is) c < mat[0].length or better yet: c < N.

Upvotes: 0

Eran
Eran

Reputation: 393821

A matrix has the same length for all rows (i.e. the same number of columns in each row), so if mat is a matrix, mat[0].length is equal to mat[i].length for any i from 0 to mat.length-1. Therefore it doesn't matter if you write mat[i].length or mat[0].length.

Upvotes: 1

Related Questions