048 Rotate Image

You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).

Example：

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]
]

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]
]

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.

解释下题目：

将矩阵按照90度进行顺时针旋转，注意不要开一个新的二维数组来处理，必须在原数组上处理

1. 利用数学方法

实际耗时：2ms

public void rotate(int[][] matrix) {
        int len = matrix.length;
        int temp = 0;

        //第一步，对角线交换
        for (int i = 0; i < len; i++) {
            for (int j = i + 1; j < len; j++) {
                temp = matrix[i][j];
                matrix[i][j] = matrix[j][i];
                matrix[j][i] = temp;
            }
        }

        //第二步，按照最中间的一列进行镜像交换
        for (int i = 0; i < len; i++) {
            for (int j = 0; j < len / 2; j++) {
                temp = matrix[i][j];
                matrix[i][j] = matrix[i][len - j - 1];
                matrix[i][len - j - 1] = temp;
            }
        }

}

  先按照对角线进行交换，然后按照把第一列和最后一列进行交换。没什么技术含量。

时间复杂度O(n^2)

空间复杂度O(1)

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2. 按照实际的交换进行交换

实际耗时：2ms

public void rotate2(int[][] matrix) {
        int n = matrix.length;
        for (int i = 0; i < n / 2; i++)
            for (int j = i; j < n - i - 1; j++) {
                int tmp = matrix[i][j];
                matrix[i][j] = matrix[n - j - 1][i];
                matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
                matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];
                matrix[j][n - i - 1] = tmp;
            }
}

  选取矩阵的四分之一，然后进行4次交换即可。

时间复杂度O(n^2)

空间复杂度O(1)

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