Subject:
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]
]
预备知识:
本题考察内容映射图像处理中的旋转操作。一张图像通过灰度处理,可以看作是由数字0-255随机组合生成的二维向量,不同的数字代表不同的灰度级,对图像的操作就等价于对矩阵的操作。
解题思路:
实现图像的90°旋转步骤如下:
1、矩阵转置(注意上三角或下三角形式在元素交换过程中的应用);
2、对称翻转(注意n为偶数和奇数时的处理方法)。
package leetcode.algroithm;
/**
* @auther lkl
* 20180203
*/
public class RotateImageBy90Degree{
public static void main(String[] lkl) {
int[][] matrix_1 = { { 1, 2, 3 }, { 4, 5, 6 }, { 7, 8, 9 } };
int[][] matrix_2 = { { 5, 1, 9, 11 }, { 2, 4, 8, 10 }, { 13, 3, 6, 7 }, { 15, 14, 12, 16 } };
// rotate(matrix_1);
rotate(matrix_2);
}
public static void rotate(int[][] matrix) {
// 首先遍历n*n矩阵,进行转置
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
if (i < j) {
int temp = 0;
temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
}
//对矩阵进行对称翻转
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;
}
}
for (int i = 0; i < matrix.length; i++) {
for (int j = 0; j < matrix.length; j++) {
System.out.print(matrix[i][j]);
}
}
}
}