原地旋转方阵 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：要在没有多余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，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) <- (y, N-1-x), (y, N-1-x)<- (N-1-x, N-1-y), (N- 1-x, N-1-y)<- (N-1-y, x), (N-1-y, x)<- (x, y)
- 打印矩阵。
C++
// C++ program to rotate a matrix
// by 90 degrees
#include
#define N 4
using namespace std;
void displayMatrix(
int mat[N][N]);
// An Inplace function to
// rotate a N x N matrix
// by 90 degrees in
// anti-clockwise direction
void rotateMatrix(int mat[][N])
{
// Consider all squares one by one
for (int x = 0; x < N / 2; x++) {
// Consider elements in group
// of 4 in current square
for (int y = x; y < N - x - 1; y++) {
// Store current cell in
// temp variable
int 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
void displayMatrix(int mat[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf("%2d ", mat[i][j]);
printf("\n");
}
printf("\n");
}
/* Driver program to test above functions */
int main()
{
// Test Case 1
int mat[N][N] = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 }
};
// Test Case 2
/* int mat[N][N] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
*/
// Test Case 3
/*int mat[N][N] = {
{1, 2},
{4, 5}
};*/
// displayMatrix(mat);
rotateMatrix(mat);
// Print rotated matrix
displayMatrix(mat);
return 0;
} Java
// Java program to rotate a
// matrix by 90 degrees
import java.io.*;
class GFG {
// An Inplace function to
// rotate a N x N matrix
// by 90 degrees in
// anti-clockwise direction
static void rotateMatrix(
int N, int mat[][])
{
// Consider all squares one by one
for (int x = 0; x < N / 2; x++) {
// Consider elements in group
// of 4 in current square
for (int y = x; y < N - x - 1; y++) {
// Store current cell in
// temp variable
int 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
static void displayMatrix(
int N, int mat[][])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
System.out.print(
" " + mat[i][j]);
System.out.print("\n");
}
System.out.print("\n");
}
/* Driver program to test above functions */
public static void main(String[] args)
{
int N = 4;
// Test Case 1
int mat[][] = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 }
};
// Test Case 2
/* int mat[][] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
*/
// Test Case 3
/*int mat[][] = {
{1, 2},
{4, 5}
};*/
// displayMatrix(mat);
rotateMatrix(N, mat);
// Print rotated matrix
displayMatrix(N, mat);
}
}
// This code is contributed by Prakriti GuptaPython3
# 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 saloni1297C#
// C# program to rotate a
// matrix by 90 degrees
using System;
class GFG {
// An Inplace function to
// rotate a N x N matrix
// by 90 degrees in anti-
// clockwise direction
static void rotateMatrix(int N,
int[, ] mat)
{
// Consider all
// squares one by one
for (int x = 0; x < N / 2; x++) {
// Consider elements
// in group of 4 in
// current square
for (int y = x; y < N - x - 1; y++) {
// store current cell
// in temp variable
int 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
static void displayMatrix(int N,
int[, ] mat)
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
Console.Write(" " + mat[i, j]);
Console.WriteLine();
}
Console.WriteLine();
}
// Driver Code
static public void Main()
{
int N = 4;
// Test Case 1
int[, ] mat = {
{ 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 }
};
// Test Case 2
/* int mat[][] =
{
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
*/
// Test Case 3
/*int mat[][] =
{
{1, 2},
{4, 5}
};*/
// displayMatrix(mat);
rotateMatrix(N, mat);
// Print rotated matrix
displayMatrix(N, mat);
}
}
// This code is contributed by ajitPHP
Javascript
C++
// C++ program to rotate a matrix
// by 90 degrees
#include
#define N 4
using namespace std;
void displayMatrix(int mat[N][N]);
// An Inplace function to
// rotate a N x N matrix
// by 90 degrees in
// anti-clockwise direction
void rotateMatrix(int mat[][N])
{ // REVERSE every row
for (int i = 0; i < N; i++)
reverse(mat[i], mat[i] + N);
// Performing Transpose
for (int i = 0; i < N; i++) {
for (int j = i; j < N; j++)
swap(mat[i][j], mat[j][i]);
}
}
// Function to print the matrix
void displayMatrix(int mat[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf("%2d ", mat[i][j]);
printf("\n");
}
printf("\n");
}
/* Driver program to test above functions */
int main()
{
// Test Case 1
int mat[N][N] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
// Test Case 2
/* int mat[N][N] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
*/
// Test Case 3
/*int mat[N][N] = {
{1, 2},
{4, 5}
};*/
// displayMatrix(mat);
rotateMatrix(mat);
// Print rotated matrix
displayMatrix(mat);
return 0;
} Python3
# Python program to rotate
# a matrix by 90 degrees
def rotateMatrix(mat):
# reversing the matrix
for i in range(len(mat)):
mat[i].reverse()
# make transpose of the matrix
for i in range(len(mat)):
for j in range(i, len(mat)):
# swapping mat[i][j] and mat[j][i]
mat[i][j], mat[j][i] = mat[j][i], mat[i][j]
# Function to print the matrix
def displayMatrix(mat):
for i in range(0, len(mat)):
for j in range(0, len(mat)):
print(mat[i][j], end=' ')
print()
mat = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]]
rotateMatrix(mat)
# Print rotated matrix
displayMatrix(mat)
# This code is contributed by shivambhagat02(CC).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
易于理解和应用
另一种方法：
1.反转每一行
2.执行转置
C++
// C++ program to rotate a matrix
// by 90 degrees
#include
#define N 4
using namespace std;
void displayMatrix(int mat[N][N]);
// An Inplace function to
// rotate a N x N matrix
// by 90 degrees in
// anti-clockwise direction
void rotateMatrix(int mat[][N])
{ // REVERSE every row
for (int i = 0; i < N; i++)
reverse(mat[i], mat[i] + N);
// Performing Transpose
for (int i = 0; i < N; i++) {
for (int j = i; j < N; j++)
swap(mat[i][j], mat[j][i]);
}
}
// Function to print the matrix
void displayMatrix(int mat[N][N])
{
for (int i = 0; i < N; i++) {
for (int j = 0; j < N; j++)
printf("%2d ", mat[i][j]);
printf("\n");
}
printf("\n");
}
/* Driver program to test above functions */
int main()
{
// Test Case 1
int mat[N][N] = { { 1, 2, 3, 4 },
{ 5, 6, 7, 8 },
{ 9, 10, 11, 12 },
{ 13, 14, 15, 16 } };
// Test Case 2
/* int mat[N][N] = {
{1, 2, 3},
{4, 5, 6},
{7, 8, 9}
};
*/
// Test Case 3
/*int mat[N][N] = {
{1, 2},
{4, 5}
};*/
// displayMatrix(mat);
rotateMatrix(mat);
// Print rotated matrix
displayMatrix(mat);
return 0;
}
Python3
# Python program to rotate
# a matrix by 90 degrees
def rotateMatrix(mat):
# reversing the matrix
for i in range(len(mat)):
mat[i].reverse()
# make transpose of the matrix
for i in range(len(mat)):
for j in range(i, len(mat)):
# swapping mat[i][j] and mat[j][i]
mat[i][j], mat[j][i] = mat[j][i], mat[i][j]
# Function to print the matrix
def displayMatrix(mat):
for i in range(0, len(mat)):
for j in range(0, len(mat)):
print(mat[i][j], end=' ')
print()
mat = [[1, 2, 3, 4],
[5, 6, 7, 8],
[9, 10, 11, 12],
[13, 14, 15, 16]]
rotateMatrix(mat)
# Print rotated matrix
displayMatrix(mat)
# This code is contributed by shivambhagat02(CC).
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) + 0(n*n) 其中 n 是数组的大小。
辅助空间： O(1)。因为需要一个恒定的空间