数组旋转的Java程序

编写一个函数rotate(ar[], d, n) 将大小为 n 的 arr[] 旋转 d 个元素。

将上述数组旋转 2 次将生成数组

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)

方法二（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 1，请将 arr[0] 存储在临时变量 temp 中，将 arr[1] 移动到 arr[0]，将 arr[2] 移动到 arr[1] ...最后将 temp 移动到 arr[n-1]

让我们以同样的例子 arr[] = [1, 2, 3, 4, 5, 6, 7], d = 2
将 arr[] 旋转 2 次
我们在第一次旋转后得到 [2, 3, 4, 5, 6, 7, 1]，在第二次旋转后得到 [3, 4, 5, 6, 7, 1, 2]。

Java

class RotateArray 
{
    /*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++)
            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
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}

Java

class RotateArray 
{
    /*Function to left rotate arr[] of size 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 (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] + " ");
    }
  
    /*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
    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
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!

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