LeetCode – Rotate Image (Java)

Category: Algorithms   January 23, 2013

You are given an n x n 2D matrix representing an image.

Rotate the image by 90 degrees (clockwise).

Follow up:
Could you do this in-place?

In-place Solution

By using the relation "matrix[i][j] = matrix[n-1-j][i]", we can loop through the matrix.

public void rotate(int[][] matrix) {
	int n = matrix.length;
	for (int i = 0; i < n / 2; i++) {
		for (int j = 0; j < Math.ceil(((double) n) / 2.); j++) {
			int temp = matrix[i][j];
			matrix[i][j] = matrix[n-1-j][i];
			matrix[n-1-j][i] = matrix[n-1-i][n-1-j];
			matrix[n-1-i][n-1-j] = matrix[j][n-1-i];
			matrix[j][n-1-i] = temp;
		}
	}
}

Related posts:

  1. LeetCode – Longest Increasing Path in a Matrix (Java)
  2. LeetCode – Maximal Square (Java)
  3. Rotate Array in Java
  4. Python: Convert Image to String, Convert String to Image
Category >> Algorithms  
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  • Alik Elzin

    I think `int j = 0` can be optimized to `int j = i`.

    Explanation:
    The algorithm iterates over the layers of the matrix – the `i` loop` – has `n/2` layers.
    The `j` loop iterates over the elements remained in the `i`-th layer.
    But, for example, element 1,0, was already handled in the `i=0` layer. No need to handle it again.

  • p.andrey

    The recursive version… just for fun 🙂

    public void rotate(int[][] a) {
    if (a == null || a.length <= 1)
    return;

    rotate(a, 0, a.length-1);
    }

    private void rotate(int[][] a, int start, int limit) {
    if (limit <= 0)
    return;

    for (int j = start; j < start+limit; j++) {
    int t = a[start][j];
    int jOff = j-start;
    a[start][j] = a[start+limit-jOff][start];
    a[start+limit-jOff][start] = a[start+limit][start+limit-jOff];
    a[start+limit][start+limit-jOff] = a[j][start+limit];
    a[j][start+limit] = t;
    }

    rotate (a, start+1, limit-2);
    }

  • NightCoder

    int length = matrix.length-1;
    System.out.println("len "+matrix.length);
    for (int i = 0; i <= (length)/2; i++) {
    for (int j = i; j < length-i; j++) {

    int temp = matrix[i][j];

    matrix[i][j] = matrix[length-j][i];
    matrix[length-j][i] = matrix[length-i][length-j];
    matrix[length-i][length-j] = matrix[j][length-i];
    matrix[j][length-i] = temp;
    }
    }

  • Guest

    why Math.ceil(((double) n) /2.) is used?

  • hina jain

    I think the j should be initialized to i instead of 0 in the second for loop.

  • svGuy

    transpose the matrix and swap rows

  • svGuy

    its easier to transpose the matrix first and then swap columns for clockwise rotation.

  • Ivy G

    You can replace “Math.ceil(((double) n) / 2.” with “(n+1)/2”

  • Chen Chen

    Great! Thanks for sharing.

  • my boo

    my bad, I got the axis messed up.

  • my boo

    Your algorithm counter rotate the image instead of clock wise rotate.

  • Guest

    If I is your matrix

    for i in range(n):
    j=0
    while(j<i):
    temp=a[i][j]
    a[i][j]=a[j][i]
    a[j][i]=temp
    j+=1

    print a

  • A Programmer

    how would you do it counter clockwise?

  • kinshuk chandra

    I liked the inplace solution. I have also done the same thing here – http://k2code.blogspot.in/2014/03/rotate-n-n-matrix-by-90-degrees.html