Python3 程序将方阵原地旋转90 度|设置 1
给定一个方阵,在不使用任何额外空间的情况下将其逆时针方向旋转 90 度。
例子 :
Input:
Matrix:
1 2 3
4 5 6
7 8 9
Output:
3 6 9
2 5 8
1 4 7
The given matrix is rotated by 90 degree
in anti-clockwise direction.
Input:
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Output:
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
The given matrix is rotated by 90 degree
in anti-clockwise direction.
这里已经讨论了需要额外空间的方法。
方法:要在没有任何额外空间的情况下解决问题,请将数组旋转为正方形,将矩阵划分为正方形或循环。例如,
一个 4 X 4 矩阵将有 2 个周期。第一个循环由它的第一行、最后一列、最后一行和第一列组成。第二个循环由第二行、倒数第二列、倒数第二行和第二列组成。这个想法是对于每个方形循环,以逆时针方向交换与矩阵中相应单元格相关的元素,即从上到左、从左到下、从下到右和从右到上一次,只使用一个临时变量来实现这一点。
示范:
First Cycle (Involves Red Elements)
1 2 3 4
5 6 7 8
9 10 11 12
13 14 15 16
Moving first group of four elements (First
elements of 1st row, last row, 1st column
and last column) of first cycle in counter
clockwise.
4 2 3 16
5 6 7 8
9 10 11 12
1 14 15 13
Moving next group of four elements of
first cycle in counter clockwise
4 8 3 16
5 6 7 15
2 10 11 12
1 14 9 13
Moving final group of four elements of
first cycle in counter clockwise
4 8 12 16
3 6 7 15
2 10 11 14
1 5 9 13
Second Cycle (Involves Blue Elements)
4 8 12 16
3 6 7 15
2 10 11 14
1 5 9 13
Fixing second cycle
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
算法:
- 在 N 边的矩阵中有 N/2 个正方形或循环。一次处理一个正方形。运行一个循环,一次循环遍历矩阵一个循环,即从0循环到N/2-1,循环计数器为i
- 考虑当前正方形中的 4 个元素组,一次旋转 4 个元素。因此,一个循环中此类组的数量为 N – 2*i。
- 所以在从 x 到 N - x - 1 的每个循环中运行一个循环,循环计数器为y
- 当前组中的元素是 (x, y), (y, N-1-x), (N-1-x, N-1-y), (N-1-y, x),现在旋转这 4 个元素,即 (x, y) 打印矩阵。
Python3
# Python3 program to rotate a matrix by 90 degrees
N = 4
# An Inplace function to rotate
# N x N matrix by 90 degrees in
# anti-clockwise direction
def rotateMatrix(mat):
# Consider all squares one by one
for x in range(0, int(N / 2)):
# Consider elements in group
# of 4 in current square
for y in range(x, N-x-1):
# store current cell in temp variable
temp = mat[x][y]
# move values from right to top
mat[x][y] = mat[y][N-1-x]
# move values from bottom to right
mat[y][N-1-x] = mat[N-1-x][N-1-y]
# move values from left to bottom
mat[N-1-x][N-1-y] = mat[N-1-y][x]
# assign temp to left
mat[N-1-y][x] = temp
# Function to print the matrix
def displayMatrix( mat ):
for i in range(0, N):
for j in range(0, N):
print (mat[i][j], end = ' ')
print ("")
# Driver Code
mat = [[0 for x in range(N)] for y in range(N)]
# Test case 1
mat = [ [1, 2, 3, 4 ],
[5, 6, 7, 8 ],
[9, 10, 11, 12 ],
[13, 14, 15, 16 ] ]
'''
# Test case 2
mat = [ [1, 2, 3 ],
[4, 5, 6 ],
[7, 8, 9 ] ]
# Test case 3
mat = [ [1, 2 ],
[4, 5 ] ]
'''
rotateMatrix(mat)
# Print rotated matrix
displayMatrix(mat)
# This code is contributed by saloni1297
输出 :
4 8 12 16
3 7 11 15
2 6 10 14
1 5 9 13
复杂性分析:
- 时间复杂度: O(n*n),其中 n 是数组的边。
需要对矩阵进行一次遍历。 - 空间复杂度: O(1)。
因为需要一个恒定的空间
请参考完整的文章 Inplace 将方阵旋转 90 度 |设置 1 了解更多详情!