给定一个维度为N 的方阵mat[][]和一个整数K ，任务是将矩阵旋转 90 度K次而不改变对角线元素的位置。
例子：
Input: mat[][] = {{1, 2, 3, 4, 5}, {6, 7, 8, 9, 10}, {11, 12, 13, 14, 15}, {16, 17, 18, 19, 20}, {21, 22, 23, 24, 25}}, K = 1
Output:
1 16 11 6 5
22 7 12 9 2
23 18 13 8 3
24 17 14 19 4
21 20 15 10 25
Input: mat[][] = {{10, 11}, {12, 13}}, K = 2
Output:
10 11
12 13
方法：利用本文讨论的思想和矩阵顺时针旋转4次后恢复的事实可以解决给定的问题。按照以下步骤解决给定的问题：
- 将K的值更新为K % 4 。
- 迭代直到K为正数并执行以下步骤：
- 穿过所述基质，对于i超过在范围[0的范围[0，N / 2）和j，N – I – 1），并执行以下步骤：
- 如果i != j和(i + j) != (N – 1) 的值，则执行以下步骤：
- 将mat[i][j]的值存储在临时变量temp 中。
- 将mat[i][j]的值更新为mat[N – 1 – j][i] 。
- 将mat[N – 1 – j][i]的值更新为mat[N – 1 -i][N – 1 – j] 。
- 将mat[N – 1 – i][N – 1 – j]的值更新为mat[j][N – 1 – i] 。
- 将mat[j][N – 1 – i]的值更新为temp 。
- 完成上述步骤后，打印得到的更新矩阵。
下面是上述方法的实现：
C++
// C++ program for the above approach
#include
using namespace std;
// Function to print the matrix
void print(vector >& mat)
{
// Iterate over the rows
for (int i = 0; i < mat.size(); i++) {
// Iterate over the columns
for (int j = 0; j < mat[0].size(); j++)
// Print the value
cout << setw(3) << mat[i][j];
cout << "\n";
}
}
// Function to perform the swapping of
// matrix elements in clockwise manner
void performSwap(vector >& mat,
int i, int j)
{
int N = mat.size();
// Stores the last row
int ei = N - 1 - i;
// Stores the last column
int ej = N - 1 - j;
// Perform the swaps
int temp = mat[i][j];
mat[i][j] = mat[ej][i];
mat[ej][i] = mat[ei][ej];
mat[ei][ej] = mat[j][ei];
mat[j][ei] = temp;
}
// Function to rotate non - diagonal
// elements of the matrix K times in
// clockwise direction
void rotate(vector >& mat,
int N, int K)
{
// Update K to K % 4
K = K % 4;
// Iterate until K is positive
while (K--) {
// Iterate each up to N/2-th row
for (int i = 0; i < N / 2; i++) {
// Iterate each column
// from i to N - i - 1
for (int j = i;
j < N - i - 1; j++) {
// Check if the element
// at i, j is not a
// diagonal element
if (i != j
&& (i + j) != N - 1) {
// Perform the swapping
performSwap(mat, i, j);
}
}
}
}
// Print the matrix
print(mat);
}
// Driver Code
int main()
{
int K = 5;
vector > mat = {
{ 1, 2, 3, 4 },
{ 6, 7, 8, 9 },
{ 11, 12, 13, 14 },
{ 16, 17, 18, 19 },
};
int N = mat.size();
rotate(mat, N, K);
return 0;
} Java
// Java program for the above approach
import java.io.*;
import java.lang.*;
import java.util.*;
public class GFG {
// Function to print the matrix
static void print(int mat[][])
{
// Iterate over the rows
for (int i = 0; i < mat.length; i++) {
// Iterate over the columns
for (int j = 0; j < mat[0].length; j++)
// Print the value
System.out.print(mat[i][j] + " ");
System.out.println();
}
}
// Function to perform the swapping of
// matrix elements in clockwise manner
static void performSwap(int mat[][], int i, int j)
{
int N = mat.length;
// Stores the last row
int ei = N - 1 - i;
// Stores the last column
int ej = N - 1 - j;
// Perform the swaps
int temp = mat[i][j];
mat[i][j] = mat[ej][i];
mat[ej][i] = mat[ei][ej];
mat[ei][ej] = mat[j][ei];
mat[j][ei] = temp;
}
// Function to rotate non - diagonal
// elements of the matrix K times in
// clockwise direction
static void rotate(int mat[][], int N, int K)
{
// Update K to K % 4
K = K % 4;
// Iterate until K is positive
while (K-- > 0) {
// Iterate each up to N/2-th row
for (int i = 0; i < N / 2; i++) {
// Iterate each column
// from i to N - i - 1
for (int j = i; j < N - i - 1; j++) {
// Check if the element
// at i, j is not a
// diagonal element
if (i != j && (i + j) != N - 1) {
// Perform the swapping
performSwap(mat, i, j);
}
}
}
}
// Print the matrix
print(mat);
}
// Driver Code
public static void main(String[] args)
{
int K = 5;
int mat[][] = {
{ 1, 2, 3, 4 },
{ 6, 7, 8, 9 },
{ 11, 12, 13, 14 },
{ 16, 17, 18, 19 },
};
int N = mat.length;
rotate(mat, N, K);
}
}
// This code is contributed by Kingash.Python3
# Python3 program for the above approach
# Function to print the matrix
def printMat(mat):
# Iterate over the rows
for i in range(len(mat)):
# Iterate over the columns
for j in range(len(mat[0])):
# Print the value
print(mat[i][j], end = " ")
print()
# Function to perform the swapping of
# matrix elements in clockwise manner
def performSwap(mat, i, j):
N = len(mat)
# Stores the last row
ei = N - 1 - i
# Stores the last column
ej = N - 1 - j
# Perform the swaps
temp = mat[i][j]
mat[i][j] = mat[ej][i]
mat[ej][i] = mat[ei][ej]
mat[ei][ej] = mat[j][ei]
mat[j][ei] = temp
# Function to rotate non - diagonal
# elements of the matrix K times in
# clockwise direction
def rotate(mat, N, K):
# Update K to K % 4
K = K % 4
# Iterate until K is positive
while (K > 0):
# Iterate each up to N/2-th row
for i in range(int(N / 2)):
# Iterate each column
# from i to N - i - 1
for j in range(i, N - i - 1):
# Check if the element
# at i, j is not a
# diagonal element
if (i != j and (i + j) != N - 1):
# Perform the swapping
performSwap(mat, i, j)
K -= 1
# Print the matrix
printMat(mat)
# Driver Code
K = 5
mat = [ [ 1, 2, 3, 4 ],
[ 6, 7, 8, 9 ],
[ 11, 12, 13, 14 ],
[ 16, 17, 18, 19 ] ]
N = len(mat)
rotate(mat, N, K)
# This code is contributed by Dharanendra L V.C#
// C# program for the above approach
using System;
public class GFG {
// Function to print the matrix
static void print(int[, ] mat)
{
// Iterate over the rows
for (int i = 0; i < mat.GetLength(0); i++) {
// Iterate over the columns
for (int j = 0; j < mat.GetLength(1); j++)
// Print the value
Console.Write(mat[i, j] + " ");
Console.WriteLine();
}
}
// Function to perform the swapping of
// matrix elements in clockwise manner
static void performSwap(int[, ] mat, int i, int j)
{
int N = mat.GetLength(0);
// Stores the last row
int ei = N - 1 - i;
// Stores the last column
int ej = N - 1 - j;
// Perform the swaps
int temp = mat[i, j];
mat[i, j] = mat[ej, i];
mat[ej, i] = mat[ei, ej];
mat[ei, ej] = mat[j, ei];
mat[j, ei] = temp;
}
// Function to rotate non - diagonal
// elements of the matrix K times in
// clockwise direction
static void rotate(int[, ] mat, int N, int K)
{
// Update K to K % 4
K = K % 4;
// Iterate until K is positive
while (K-- > 0) {
// Iterate each up to N/2-th row
for (int i = 0; i < N / 2; i++) {
// Iterate each column
// from i to N - i - 1
for (int j = i; j < N - i - 1; j++) {
// Check if the element
// at i, j is not a
// diagonal element
if (i != j && (i + j) != N - 1) {
// Perform the swapping
performSwap(mat, i, j);
}
}
}
}
// Print the matrix
print(mat);
}
// Driver Code
public static void Main(string[] args)
{
int K = 5;
int[, ] mat = {
{ 1, 2, 3, 4 },
{ 6, 7, 8, 9 },
{ 11, 12, 13, 14 },
{ 16, 17, 18, 19 },
};
int N = mat.GetLength(0);
rotate(mat, N, K);
}
}
// This code is contributed by ukasp.Javascript
输出：
1 11 6 4
17 7 8 2
18 12 13 3
16 14 9 19时间复杂度： O(N 2 )
辅助空间： O(1)
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