给定一幅由N × N矩阵表示的图像,其中每个像素的大小为4字节,编写一种方法,将图像旋转90度。
不占用额外内存空间能否做到?
示例 1:
给定 matrix = [ [1,2,3], [4,5,6], [7,8,9] ], 原地旋转输入矩阵,使其变为: [ [7,4,1], [8,5,2], [9,6,3] ]
示例 2:
给定 matrix = [ [ 5, 1, 9,11], [ 2, 4, 8,10], [13, 3, 6, 7], [15,14,12,16] ], 原地旋转输入矩阵,使其变为: [ [15,13, 2, 5], [14, 3, 4, 1], [12, 6, 8, 9], [16, 7,10,11] ]
原地旋转,i 的范围是 [0, n/2)
,j 的范围是 [i, n-1-i)
。
class Solution:
def rotate(self, matrix: List[List[int]]) -> None:
"""
Do not return anything, modify matrix in-place instead.
"""
n = len(matrix)
for i in range(n // 2):
for j in range(i, n - 1 - i):
t = 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] = t
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 t = 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] = t;
}
}
}
}