LeetCode 48. Rotate Image

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

Rotate the image by 90 degrees (clockwise).

Note:
You have to rotate the image in-place, which means you have to modify the input 2D matrix directly. DO NOT allocate another 2D matrix and do the rotation.

Example 1:

Given input matrix = 
[
  [1,2,3],
  [4,5,6],
  [7,8,9]
],

rotate the input matrix in-place such that it becomes:
[
  [7,4,1],
  [8,5,2],
  [9,6,3]
]

Example 2:

Given input matrix =
[
  [ 5, 1, 9,11],
  [ 2, 4, 8,10],
  [13, 3, 6, 7],
  [15,14,12,16]
], 

rotate the input matrix in-place such that it becomes:
[
  [15,13, 2, 5],
  [14, 3, 4, 1],
  [12, 6, 8, 9],
  [16, 7,10,11]
]

我刚开始想到的是比较暴力的算法，即把所有数字都用list保存起来，然后再让他们回到各自的位置：

public void rotate(int[][] matrix) {
        int length = matrix.length;
        List<Integer> list = new ArrayList<>();

        for (int i = 0; i < length; i++) {
            for (int j = 0; j < length; j++) {
                list.add(matrix[i][j]);
            }
        }

        for (int i = 0; i < length; i++) {
            for (int j = length - 1; j >= 0; j--) {
                matrix[j][i] = list.remove(list.size() - 1);
            }
        }
    }

还有一种思路是，四个位置轮流交换：

思路图

思路就是，四个位置相互交换一下就可以。

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