[leedcode 48] Rotate Image

標籤：

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) {        //本題一定要注意邊界。每一層迴圈代表反轉一層（從外到內迴圈），一共有len/2層。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) {        //本題的另一種解法，首先沿著水平中線翻轉一次，然後沿著主對角線，翻轉一次，最終能夠實現順時針旋轉90度        //或者，首先沿著副對角線翻轉一次，然後沿著水平中線翻轉一次。        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;                         }        }             }}

 

[leedcode 48] Rotate Image