算法48. Rotate Image

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]
]

class Solution {
    public void rotate(int[][] matrix) {
        
    }
}

给一个 n * n 的 2D 矩阵表示图片，顺时针旋转90。
Note： 要直接修改 2D 矩阵，不能使用额外的 2D 矩阵。

解：
2D 矩阵，结构上表示为一个二维数组，如果要直接旋转不太好操作。对于矩阵的操作，顺时针旋转90度，等于左下部与右上部交换，再左半部与右半部交换。即先交换matrix[i][j]与matrix[j][i]，再对每一行进行matrix[i][j]与 matrix[i][matrix.length-1-j]的交换。代码如下：

public void rotate(int[][] matrix) {
    // 先把左下与右上进行交换
    for (int i = 0; i < matrix.length; i++) {
        for (int j = i; j < matrix[0].length; j++) {
            int temp = 0;
            temp = matrix[i][j];
            matrix[i][j] = matrix[j][i];
            matrix[j][i] = temp;
        }
    }
    // 再对每一行进行对位交换
    for (int i = 0; i < matrix.length; i++) {
        for (int j = 0; j < matrix.length / 2; j++) {
            int temp = 0;
            temp = matrix[i][j];
            matrix[i][j] = matrix[i][matrix.length - 1 - j];
            matrix[i][matrix.length - 1 - j] = temp;
        }
    }
}