48. 旋转图像
给定一个 n × n 的二维矩阵 matrix
表示一个图像。请你将图像顺时针旋转 90 度。
你必须在 原地 旋转图像,这意味着你需要直接修改输入的二维矩阵。请不要 使用另一个矩阵来旋转图像。
示例 1:
输入:matrix = [[1,2,3],[4,5,6],[7,8,9]]
输出:[[7,4,1],[8,5,2],[9,6,3]]
示例 2:
输入:matrix = [[5,1,9,11],[2,4,8,10],[13,3,6,7],[15,14,12,16]]
输出:[[15,13,2,5],[14,3,4,1],[12,6,8,9],[16,7,10,11]]
示例 3:
输入:matrix = [[1]]
输出:[[1]]
示例 4:
输入:matrix = [[1,2],[3,4]]
输出:[[3,1],[4,2]]
提示:
matrix.length == n
matrix[i].length == n
1 <= n <= 20
-1000 <= matrix[i][j] <= 1000
解题思路
方形矩阵旋转公式
[left{egin{array}{ll}
ext { temp } & =operatorname{matrix}[operatorname{row}][operatorname{col}] \
ext { matrix }[ ext { row }][operatorname{col}] & =operatorname{matrix}[n-operatorname{col}-1][operatorname{row}] \
ext { matrix }[n-operatorname{col}-1][ ext { row }] & =operatorname{matrix}[n-operatorname{row}-1][n-operatorname{col}-1] 2 \
ext { matrix }[n-operatorname{row}-1][n-operatorname{col}-1] & =operatorname{matrix}[operatorname{col}[n- ext { row }-1] \
operatorname{matrix}[operatorname{col}[n-operatorname{row}-1] & =t e m p
end{array}
ight.
]
public void rotate(int[][] matrix) {
int n = matrix.length;
for (int i = 0; i < n / 2; ++i) {
for (int j = 0; j < (n + 1) / 2; ++j) {
int temp = 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] = temp;
}
}
}