# LeetCode – Rotate Image (Java)

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

Rotate the image by 90 degrees (clockwise).

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; } } }```
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

• Guest

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