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

給你一個n*n表示一張圖片的2維矩陣。

Rotate the image by 90 degrees (clockwise).

按照順時針方向旋轉90°。

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.

你只能直接利用矩陣旋轉，意思是只能直接修改傳入的2維矩陣，不要重新構造一個2維矩陣來進行旋轉。

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]
]

解：

我們可以先按照對角線對調整個矩陣，然后再按行做對調，即可完成順時針90°的旋轉

    public static void rotate(int[][] matrix) {
        int n = matrix.length;
        //沿對角線對折，右上角變?yōu)樽笙陆?        for(int i = 0; i<n;i++) {
            for(int j = i; j<n;j++) {//j=i是要從每個小矩形左上角開始
                int tmp = matrix[j][i];
                matrix[j][i] = matrix[i][j];
                matrix[i][j] = tmp;
            }
        }
        //再反轉每一行 就相當于原始矩陣轉了90度
        for(int i =0;i<n;i++) {     //每一行都需要旋轉
            for(int j = 0; j<n/2 ; j++) { //一行中只需要遍歷一半，因為另一半已經(jīng)被對調了
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[i][n-1-j];
                matrix[i][n-1-j] = tmp;
            }
        }
    }