You are given an n x n 2D matrix representing an image.
给你一个n*n表示一张图片的2维矩阵。
Rotate the image by 90 degrees (clockwise).
按照顺时针方向旋转90°。
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.
你只能直接利用矩阵旋转，意思是只能直接修改传入的2维矩阵，不要重新构造一个2维矩阵来进行旋转。
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]
]
解：
我们可以先按照对角线对调整个矩阵，然后再按行做对调，即可完成顺时针90°的旋转
public static void rotate(int[][] matrix) {
int n = matrix.length;
//沿对角线对折，右上角变为左下角
for(int i = 0; i<n;i++) {
for(int j = i; j<n;j++) {//j=i是要从每个小矩形左上角开始
int tmp = matrix[j][i];
matrix[j][i] = matrix[i][j];
matrix[i][j] = tmp;
}
}
//再反转每一行 就相当于原始矩阵转了90度
for(int i =0;i<n;i++) { //每一行都需要旋转
for(int j = 0; j<n/2 ; j++) { //一行中只需要遍历一半，因为另一半已经被对调了
int tmp = matrix[i][j];
matrix[i][j] = matrix[i][n-1-j];
matrix[i][n-1-j] = tmp;
}
}
}
