You are given an n x n 2D matrix representing an image.
Rotate the image by 90 degrees (clockwise).
Example：
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]
]
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]
]
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.
解释下题目：
将矩阵按照90度进行顺时针旋转，注意不要开一个新的二维数组来处理，必须在原数组上处理
1. 利用数学方法
实际耗时：2ms
public void rotate(int[][] matrix) {
int len = matrix.length;
int temp = 0;
//第一步，对角线交换
for (int i = 0; i < len; i++) {
for (int j = i + 1; j < len; j++) {
temp = matrix[i][j];
matrix[i][j] = matrix[j][i];
matrix[j][i] = temp;
}
}
//第二步，按照最中间的一列进行镜像交换
for (int i = 0; i < len; i++) {
for (int j = 0; j < len / 2; j++) {
temp = matrix[i][j];
matrix[i][j] = matrix[i][len - j - 1];
matrix[i][len - j - 1] = temp;
}
}
}
先按照对角线进行交换，然后按照把第一列和最后一列进行交换。没什么技术含量。
时间复杂度O(n^2)
空间复杂度O(1)
2. 按照实际的交换进行交换
实际耗时：2ms
public void rotate2(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < n / 2; i++)
for (int j = i; j < n - i - 1; j++) {
int tmp = matrix[i][j];
matrix[i][j] = matrix[n - j - 1][i];
matrix[n - j - 1][i] = matrix[n - i - 1][n - j - 1];
matrix[n - i - 1][n - j - 1] = matrix[j][n - i - 1];
matrix[j][n - i - 1] = tmp;
}
}
选取矩阵的四分之一，然后进行4次交换即可。
