用于将矩阵旋转 45 度的 Python3 程序
给定一个大小为N*N的矩阵mat[][] ,任务是将矩阵旋转45 度并打印矩阵。
例子:
Input: N = 6,
mat[][] = {{3, 4, 5, 1, 5, 9, 5},
{6, 9, 8, 7, 2, 5, 2},
{1, 5, 9, 7, 5, 3, 2},
{4, 7, 8, 9, 3, 5, 2},
{4, 5, 2, 9, 5, 6, 2},
{4, 5, 7, 2, 9, 8, 3}}
Output:
3
6 4
1 9 5
4 5 8 1
4 7 9 7 5
4 5 8 7 2 9
5 2 9 5 5
7 9 3 3
2 5 5
9 6
8
Input: N = 4,
mat[][] = {{2, 5, 7, 2},
{9, 1, 4, 3},
{5, 8, 2, 3},
{6, 4, 6, 3}}
Output:
2
9 5
5 1 7
6 8 4 2
4 2 3
6 3
3
方法:按照以下步骤解决问题:
- 使用计数器变量将对角线元素存储在列表中。
- 打印使输出看起来像所需模式所需的空格数。
- 反转列表后打印列表元素。
- 仅遍历对角元素以优化操作所花费的时间。
下面是上述方法的实现:
Python3
# Python3 program for the above approach
# Function to rotate matrix by 45 degree
def matrix(n, m, li):
# Counter Variable
ctr = 0
while(ctr < 2 * n-1):
print(" "*abs(n-ctr-1), end ="")
lst = []
# Iterate [0, m]
for i in range(m):
# Iterate [0, n]
for j in range(n):
# Diagonal Elements
# Condition
if i + j == ctr:
# Appending the
# Diagonal Elements
lst.append(li[i][j])
# Printing reversed Diagonal
# Elements
lst.reverse()
print(*lst)
ctr += 1
# Driver Code
# Dimensions of Matrix
n = 8
m = n
# Given matrix
li = [[4, 5, 6, 9, 8, 7, 1, 4],
[1, 5, 9, 7, 5, 3, 1, 6],
[7, 5, 3, 1, 5, 9, 8, 0],
[6, 5, 4, 7, 8, 9, 3, 7],
[3, 5, 6, 4, 8, 9, 2, 1],
[3, 1, 6, 4, 7, 9, 5, 0],
[8, 0, 7, 2, 3, 1, 0, 8],
[7, 5, 3, 1, 5, 9, 8, 5]]
# Function Call
matrix(n, m, li)
4
1 5
7 5 6
6 5 9 9
3 5 3 7 8
3 5 4 1 5 7
8 1 6 7 5 3 1
7 0 6 4 8 9 1 4
5 7 4 8 9 8 6
3 2 7 9 3 0
1 3 9 2 7
5 1 5 1
9 0 0
8 8
5
时间复杂度: O(N 2 )
辅助空间: O(1)
有关详细信息,请参阅有关将矩阵旋转 45 度的完整文章!