05 August 2008
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:
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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:
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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]
]
计算机里图片的本质是矩阵，旋转矩阵即是旋转图片，有很多方法可以旋转矩阵，我自己比较好理解的两种办法是：
1）首先对原数组取其转置矩阵（行列互换），然后把每行的数字翻转可得到结果，如下所示(其中蓝色数字表示翻转轴)：
1 2 3 1 4 7 7 4 1
4 5 6 –> 2 5 8 –> 8 5 2
7 8 9 3 6 9 9 6 3
2）首先以从对角线为轴翻转，然后再以x轴中线上下翻转即可得到结果：
1 2 3 9 6 3 7 4 1
4 5 6 –> 8 5 2 –> 8 5 2
7 8 9 7 4 1 9 6 3
3）每次循环换四个数字：
1 2 3 7 2 1 7 4 1
4 5 6 –> 4 5 6 –> 8 5 2
7 8 9 9 8 3 9 6 3
转置矩阵的办法
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class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < n; i++) { // 将右上角和左下角交换
for (int j = i; j < n; j++) {//j = i不用重复转置
//转换为转置矩阵transport matrix
int temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
//逐行将元素翻转
for (int i = 0; i < n; i++) {
for (int j = 0; j < n/2; j++) {//注意这里是j < n/2，没有=
int temp = matrix[i][j];
matrix[i][j] = matrix[i][n - j - 1];
matrix[i][n - j - 1] = temp;
}
}
}
}
对角线翻转的办法
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class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
//以对角线为轴翻转
for (int i = 0; i < n - 1; i++) {
for (int j = 0; j < n - i; j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[n - 1 - j][n - 1 - i];
matrix[n - 1 - j][n - 1 - i] = temp;
}
}
//以x轴中线上下翻转
for (int i = 0; i < n / 2; i++) {
for (int j = 0; j < n; j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[n - 1 - i][j];
matrix[n - 1 - i][j] = temp;
}
}
}
}
直接旋转
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class Solution {
public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < n / 2; i++) {
for (int j = i; j < n - 1 - i; j++) {
int temp = matrix[i][j];
matrix[i][j] = matrix[n - 1 - j][i];
matrix[n - 1 - j][i] = matrix[n - 1 - i][n - 1 - j];
matrix[n - 1 - i][n - 1 - j] = matrix[j][n - 1 - i];
matrix[j][n - 1 - i] = temp;
}
}
}
}