数组旋转程序

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 the first 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]。
下面是上述方法的实现：

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], i;
    for (i = 0; i < n - 1; i++)
        arr[i] = arr[i + 1];
 
    arr[n-1] = 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 n)
{
    for (int i = 0; i < n; 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

// C program to rotate an array by
// d elements
#include 
 
/* 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 temp = arr[0], i;
    for (i = 0; i < n - 1; i++)
        arr[i] = arr[i + 1];
    arr[n-1] = temp;
}
 
/* utility function to print an array */
void printArray(int arr[], int n)
{
    int i;
    for (i = 0; i < n; 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);
    return 0;
}

Java

// Java program to rotate an array by
// d elements
 
class RotateArray {
    /*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);
    }
 
    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[n-1] = temp;
    }
 
    /* utility function to print an array */
    void printArray(int arr[], int n)
    {
        for (int i = 0; i < n; i++)
            System.out.print(arr[i] + " ");
    }
 
    // Driver program to test above functions
    public static void main(String[] args)
    {
        RotateArray rotate = new RotateArray();
        int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
        rotate.leftRotate(arr, 2, 7);
        rotate.printArray(arr, 7);
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3

# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d*/
def leftRotate(arr, d, n):
    for i in range(d):
        leftRotatebyOne(arr, n)
 
# Function to left Rotate arr[] of size n by 1*/
def leftRotatebyOne(arr, n):
    temp = arr[0]
    for i in range(n-1):
        arr[i] = arr[i + 1]
    arr[n-1] = temp
         
 
# utility function to print an array */
def printArray(arr, size):
    for i in range(size):
        print ("% d"% arr[i], end =" ")
 
  
# Driver program to test above functions */
arr = [1, 2, 3, 4, 5, 6, 7]
leftRotate(arr, 2, 7)
printArray(arr, 7)
 
# This code is contributed by Shreyanshi Arun

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[n-1] = 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

Javascript

 
// JavaScript program to rotate an array by
// d elements
 
 
    /* Function to left rotate arr of size n by d */
    function leftRotate(arr , d , n) {
        for (i = 0; i < d; i++)
            leftRotatebyOne(arr, n);
    }
 
    function leftRotatebyOne(arr , n) {
        var i, temp;
        temp = arr[0];
        for (i = 0; i < n - 1; i++)
            arr[i] = arr[i + 1];
        arr[n - 1] = temp;
    }
 
    /* utility function to print an array */
    function printArray(arr , n) {
        for (i = 0; i < n; i++)
            document.write(arr[i] + " ");
    }
 
    // Driver program to test above functions
     
        var arr = [ 1, 2, 3, 4, 5, 6, 7 ];
        leftRotate(arr, 2, 7);
        printArray(arr, 7);
 
// This code is contributed by todaysgaurav
 

输出 ：

3 4 5 6 7 1 2 

时间复杂度： O(n * d)
辅助空间： O(1)
方法3(一种杂耍算法)
这是方法2的扩展。不要将数组一一移动，而是将数组分成不同的集合
其中集合数等于n和d的GCD并在集合内移动元素。
如果对于上面的示例数组(n = 7和d = 2)，GCD为1，则元素将仅在一组内移动，我们仅从temp = arr [0]开始并继续移动arr [I + d]到arr [I]，最后将温度存储在正确的位置。
这是n = 12且d = 3的示例。GCD为3且

Let 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}

下面是上述方法的实现：

C++

// C++ program to rotate an array by
// d elements
#include 
using namespace std;
 
/*Fuction 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)
{
    /* To handle if d >= n */
    d = d % n;
    int g_c_d = gcd(d, n);
    for (int i = 0; i < g_c_d; 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

// C program to rotate an array by
// d elements
#include 
 
/* function to print an array */
void printArray(int arr[], int size);
 
/*Fuction 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;
    /* To handle if d >= n */
    d = d % n;
    int g_c_d = gcd(d, n);
    for (i = 0; i < g_c_d; 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 n)
{
    int i;
    for (i = 0; i < n; i++)
        printf("%d ", arr[i]);
}
 
/*Fuction 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;
}

Java

// Java program to rotate an array by
// d elements
class RotateArray {
    /*Function to left rotate arr[] of siz n by d*/
    void leftRotate(int arr[], int d, int n)
    {
        /* To handle if d >= n */
        d = d % n;
        int i, j, k, temp;
        int g_c_d = gcd(d, n);
        for (i = 0; i < g_c_d; 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 */
    void printArray(int arr[], int size)
    {
        int i;
        for (i = 0; i < size; i++)
            System.out.print(arr[i] + " ");
    }
 
    /*Fuction 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
    public static void main(String[] args)
    {
        RotateArray rotate = new RotateArray();
        int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
        rotate.leftRotate(arr, 2, 7);
        rotate.printArray(arr, 7);
    }
}
 
// This code has been contributed by Mayank Jaiswal

Python3

# Python3 program to rotate an array by
# d elements
# Function to left rotate arr[] of size n by d
def leftRotate(arr, d, n):
    d = d % n
    g_c_d = gcd(d, n)
    for i in range(g_c_d):
         
        # 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
def printArray(arr, size):
    for i in range(size):
        print ("% d" % arr[i], end =" ")
 
# Fuction to get gcd of a and b
def gcd(a, b):
    if b == 0:
        return a;
    else:
        return gcd(b, a % b)
 
# Driver program to test above functions
arr = [1, 2, 3, 4, 5, 6, 7]
n = len(arr)
d = 2
leftRotate(arr, d, n)
printArray(arr, n)
 
# This code is contributed by Shreyanshi Arun

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;
        /* To handle if d >= n */
        d = d % n;
        int g_c_d = gcd(d, n);
        for (i = 0; i < g_c_d; 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] + " ");
    }
 
    /* Fuction 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

Java脚本

 
// JavaScript program to rotate an array by
// d elements
 
/*Fuction to get gcd of a and b*/
function gcd( a, b){
    if (b == 0)
        return a;
    else
        return gcd(b, a % b);
}
 
/*Function to left rotate arr[] of siz n by d*/
function leftRotate(arr, d, n){
    /* To handle if d >= n */
    d = d % n;
    let g_c_d = gcd(d, n);
    for (let i = 0; i < g_c_d; i++) {
        /* move i-th values of blocks */
        let temp = arr[i];
        let j = i;
 
        while (1) {
            let 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
function printArray(arr, size){
    for (let i = 0; i < size; i++)
        document.write(arr[i] +" ");
}
 
/* Driver program to test above functions */
let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
let n = arr.length;
// Function calling
leftRotate(arr, 2, n);
printArray(arr, n);
 

输出 ：

3 4 5 6 7 1 2 

时间复杂度： O(n)
辅助空间： O(1)