LeetCode-48~Rotate Image

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

• 将原始矩阵中的元素复制到新矩阵中
• 矩阵中元素位置计算：横纵坐标分别计算
  • 原始矩阵中横坐标为新矩阵中纵坐标
  • 原始矩阵中纵坐标j为新矩阵中height-j-1坐标

Java代码

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：

原地坐标转换。如图所示：

原地坐标转换

Java代码

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;
        
        //原地坐标转换
        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;
            }
        }
    }
}

参考

• LeetCode
• 开源中国oschina