编写一个函数turn(ar []，d，n)，该函数将大小为n的arr []旋转d个元素。 
将上面的数组旋转2将使数组
方法1(使用临时数组)
Input arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2, n =7
1) Store d elements in a temp array
temp[] = [1, 2]
2) Shift rest of the arr[]
arr[] = [3, 4, 5, 6, 7, 6, 7]
3) Store back the d elements
arr[] = [3, 4, 5, 6, 7, 1, 2]时间复杂度： O(n)
辅助空间： O(d)
方法2(一一旋转)
leftRotate(arr[], d, n)
start
For i = 0 to i < d
Left rotate all elements of arr[] by one
end要旋转一个，将arr [0]存储在一个临时变量temp中，将arr [1]移至arr [0]，将arr [2]移至arr [1]…最后将temp移至arr [n-1]
让我们以相同的示例arr [] = [1、2、3、4、5、6、7]，d = 2
将arr []旋转1倍
第一次旋转后得到[2，3，4，5，5，6，7，1]，第二次旋转后得到[3，4，5，6，7，1，2]。
C++
// C++ program to rotate an array by
// d elements
#include
using namespace std;
/*Function to left Rotate arr[] of
size n by 1*/
void leftRotatebyOne(int arr[], int n)
{
int temp = arr[0];
int i;
for(i = 0; i < n-1; i++)
arr[i] = arr[i+1];
arr[i] = temp;
}
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
/* utility function to print an array */
void printArray(int arr[], int size)
{
for(int i = 0; i < size; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
} C
/*Function to left Rotate arr[] of size n by 1*/
void leftRotatebyOne(int arr[], int n);
/*Function to left rotate arr[] of size n by d*/
void leftRotate(int arr[], int d, int n)
{
int i;
for (i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
void leftRotatebyOne(int arr[], int n)
{
int i, temp;
temp = arr[0];
for (i = 0; i < n-1; i++)
arr[i] = arr[i+1];
arr[i] = temp;
}
/* utility function to print an array */
void printArray(int arr[], int size)
{
int i;
for(i = 0; i < size; i++)
printf("%d ", arr[i]);
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
leftRotate(arr, 2, 7);
printArray(arr, 7);
getchar();
return 0;
}C#
// C# program for array rotation
using System;
class GFG
{
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int []arr, int d,
int n)
{
for (int i = 0; i < d; i++)
leftRotatebyOne(arr, n);
}
static void leftRotatebyOne(int []arr, int n)
{
int i, temp = arr[0];
for (i = 0; i < n - 1; i++)
arr[i] = arr[i + 1];
arr[i] = temp;
}
/* utility function to print an array */
static void printArray(int []arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 3, 4, 5, 6, 7};
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
// This code is contributed by Sam007
Output :3 4 5 6 7 1 2
Time complexity : O(n * d)Auxiliary Space : O(1)METHOD 3 (A Juggling Algorithm)This is an extension of method 2. Instead of moving one by one, divide the array in different setswhere number of sets is equal to GCD of n and d and move the elements within sets.If GCD is 1 as is for the above example array (n = 7 and d =2), then elements will be moved within one set only, we just start with temp = arr[0] and keep moving arr[I+d] to arr[I] and finally store temp at the right place.Here is an example for n =12 and d = 3. GCD is 3 andLet arr[] be {1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11, 12}
a) Elements are first moved in first set – (See below diagram for this movement)
arr[] after this step --> {4 2 3 7 5 6 10 8 9 1 11 12}
b) Then in second set.
arr[] after this step --> {4 5 3 7 8 6 10 11 9 1 2 12}
c) Finally in third set.
arr[] after this step --> {4 5 6 7 8 9 10 11 12 1 2 3}C++
// C++ program to rotate an array by
// d elements
#include
using namespace std;
/*Function to get gcd of a and b*/
int gcd(int a,int b)
{
if(b == 0)
return a;
else
return gcd(b, a%b);
}
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
for (int i = 0; i < gcd(d, n); i++)
{
/* move i-th values of blocks */
int temp = arr[i];
int j = i;
while(1)
{
int k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
// Function to print an array
void printArray(int arr[], int size)
{
for(int i = 0; i < size; i++)
cout << arr[i] << " ";
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
int n = sizeof(arr) / sizeof(arr[0]);
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
return 0;
} C
/* function to print an array */
void printArray(int arr[], int size);
/*Function to get gcd of a and b*/
int gcd(int a,int b);
/*Function to left rotate arr[] of siz n by d*/
void leftRotate(int arr[], int d, int n)
{
int i, j, k, temp;
for (i = 0; i < gcd(d, n); i++)
{
/* move i-th values of blocks */
temp = arr[i];
j = i;
while(1)
{
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* function to print an array */
void printArray(int arr[], int size)
{
int i;
for(i = 0; i < size; i++)
printf("%d ", arr[i]);
}
/*Function to get gcd of a and b*/
int gcd(int a,int b)
{
if(b==0)
return a;
else
return gcd(b, a%b);
}
/* Driver program to test above functions */
int main()
{
int arr[] = {1, 2, 3, 4, 5, 6, 7};
leftRotate(arr, 2, 7);
printArray(arr, 7);
getchar();
return 0;
}C#
// C# program for array rotation
using System;
class GFG
{
/* Function to left rotate arr[]
of size n by d*/
static void leftRotate(int []arr, int d,
int n)
{
int i, j, k, temp;
for (i = 0; i < gcd(d, n); i++)
{
/* move i-th values of blocks */
temp = arr[i];
j = i;
while (true)
{
k = j + d;
if (k >= n)
k = k - n;
if (k == i)
break;
arr[j] = arr[k];
j = k;
}
arr[j] = temp;
}
}
/*UTILITY FUNCTIONS*/
/* Function to print an array */
static void printArray(int []arr, int size)
{
for (int i = 0; i < size; i++)
Console.Write(arr[i] + " ");
}
/* Function to get gcd of a and b*/
static int gcd(int a, int b)
{
if (b == 0)
return a;
else
return gcd(b, a % b);
}
// Driver code
public static void Main()
{
int []arr = {1, 2, 3, 4, 5, 6, 7};
leftRotate(arr, 2, 7);
printArray(arr, 7);
}
}
// This code is contributed by Sam007
Output :3 4 5 6 7 1 2
Time complexity : O(n)Auxiliary Space : O(1)Please refer complete article on Program for array rotation for more details!Want to learn from the best curated videos and practice problems, check out the C Foundation Course for Basic to Advanced C.