[leedcode 48] Rotate Image

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 order to fulfill the follow up requirement,  i.e. in-place, we should utilize a temporary int variable and switch the values in the matrix.  Coming back to our problem, rotating a matrix can be divided into steps and each step responses to rotating specific layer of the matrix.  For example, when n=6 there are n/2 = 3 steps from outside layers to inside layers.

public class Solution {
    /*public void rotate(int[][] matrix) {
        //本题一定要注意边界。每一层循环代表反转一层（从外到内循环），一共有len/2层。注意j的取值，j>=i,j<len-i-1;

        int len=matrix.length;
        for(int i=0;i<len/2;i++){
            for(int j=i;j<len-i-1;j++){
                int temp=matrix[i][j];
                matrix[i][j]=matrix[len-1-j][i];
                matrix[len-1-j][i]=matrix[len-1-i][len-1-j];
                matrix[len-1-i][len-1-j]=matrix[j][len-i-1];
                matrix[j][len-i-1]=temp;

            }
        }
    }*/

    public void rotate(int[][] matrix) {
        //本题的另一种解法，首先沿着水平中线翻转一次，然后沿着主对角线，翻转一次，最终能够实现顺时针旋转90度
        //或者，首先沿着副对角线翻转一次，然后沿着水平中线翻转一次。
        int len=matrix.length;
        int i=0;
        int j=0;
        for(;i<len/2;i++){
            for(j=0;j<len;j++){
                int temp=matrix[i][j];
                matrix[i][j]=matrix[len-1-i][j];
                matrix[len-1-i][j]=temp;

            }
        }
        for(i=0;i<len;i++){
            for(j=i+1;j<len;j++){
                int temp=matrix[i][j];
                matrix[i][j]=matrix[j][i];
                matrix[j][i]=temp;

            }
        } 

    }
}