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?

算法分析

方法一

• 將原始矩陣中的元素復制到新矩陣中
• 矩陣中元素位置計算：橫縱坐標分別計算
  • 原始矩陣中橫坐標為新矩陣中縱坐標
  • 原始矩陣中縱坐標j為新矩陣中height-j-1坐標

Java代碼

public class Solution {
    public void rotate(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return;
        int height = matrix.length;
        int weight = matrix[0].length;
        
        if (height != weight) return;
        
        int[][] temp = new int[weight][height];
        for (int i = 0; i < height; i ++) {
            for (int j = 0; j < weight; j ++) {
                temp[i][j] = matrix[i][j];
            }
        }

        for (int i = 0; i < height; i ++) {
            for (int j = 0; j < weight; j ++) {
                matrix[i][j] = temp[height - j- 1][i];
            }
        }    
    }
}

方法二：

原地坐標轉換。如圖所示：

原地坐標轉換

Java代碼

public class Solution {
    public void rotate(int[][] matrix) {
        if (matrix == null || matrix.length == 0 || matrix[0].length == 0) return;
        int height = matrix.length;
        int weight = matrix[0].length;
        
        if (height != weight) return;
        
        int temp = 0;
        
        //原地坐標轉換
        for (int i = 0; i < height / 2; i ++) {
            for (int j = i; j < height - 1 - i; j ++) {
                temp = matrix[i][j];
                matrix[i][j] = matrix[height - j - 1][i];
                matrix[height - j - 1][i] = matrix[height - i - 1][height - j - 1];
                matrix[height - i - 1][height - j - 1] = matrix[j][height - i - 1];
                matrix[j][height - i - 1] = temp;
            }
        }
    }
}

參考

• LeetCode
• 開源中國oschina