给定大小为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：按照下面给出的步骤来解决问题：
- 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图案所需的空格数。
- 反转列表后打印列表元素。
- 仅遍历对角线元素以优化操作所花费的时间。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to rotate matrix by 45 degree
void matrix(int n, int m, vector> li)
{
// Counter Variable
int ctr = 0;
while (ctr < 2 * n - 1)
{
for(int i = 0;
i < abs(n - ctr - 1);
i++)
{
cout << " ";
}
vector lst;
// Iterate [0, m]
for(int i = 0; i < m; i++)
{
// Iterate [0, n]
for(int j = 0; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
lst.push_back(li[i][j]);
}
}
}
// Printing reversed Diagonal
// Elements
for(int i = lst.size() - 1; i >= 0; i--)
{
cout << lst[i] << " ";
}
cout << endl;
ctr += 1;
}
}
// Driver code
int main()
{
// Dimensions of Matrix
int n = 8;
int m = n;
// Given matrix
vector> 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);
return 0;
}
// This code is contributed by divyeshrabadiya07 Java
// Java program for
// the above approach
import java.util.*;
class GFG{
// Function to rotate
// matrix by 45 degree
static void matrix(int n, int m,
int [][]li)
{
// Counter Variable
int ctr = 0;
while (ctr < 2 * n - 1)
{
for(int i = 0; i < Math.abs(n - ctr - 1);
i++)
{
System.out.print(" ");
}
Vector lst = new Vector();
// Iterate [0, m]
for(int i = 0; i < m; i++)
{
// Iterate [0, n]
for(int j = 0; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
lst.add(li[i][j]);
}
}
}
// Printing reversed Diagonal
// Elements
for(int i = lst.size() - 1; i >= 0; i--)
{
System.out.print(lst.get(i) + " ");
}
System.out.println();
ctr += 1;
}
}
// Driver code
public static void main(String[] args)
{
// Dimensions of Matrix
int n = 8;
int m = n;
// Given matrix
int[][] 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);
}
}
// This code is contributed by Princi Singh 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)C#
// C# program for
// the above approach
using System;
using System.Collections;
class GFG{
// Function to rotate
// matrix by 45 degree
static void matrix(int n, int m,
int [,]li)
{
// Counter Variable
int ctr = 0;
while (ctr < 2 * n - 1)
{
for(int i = 0;
i < Math.Abs(n - ctr - 1);
i++)
{
Console.Write(" ");
}
ArrayList lst = new ArrayList();
// Iterate [0, m]
for(int i = 0; i < m; i++)
{
// Iterate [0, n]
for(int j = 0; j < n; j++)
{
// Diagonal Elements
// Condition
if (i + j == ctr)
{
// Appending the
// Diagonal Elements
lst.Add(li[i, j]);
}
}
}
// Printing reversed Diagonal
// Elements
for(int i = lst.Count - 1;
i >= 0; i--)
{
Console.Write((int)lst[i] + " ");
}
Console.Write("\n");
ctr += 1;
}
}
// Driver code
public static void Main(string[] args)
{
// Dimensions of Matrix
int n = 8;
int m = n;
// Given matrix
int[,] 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);
}
}
// This code is contributed by Rutvik_56输出：
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)
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