算法：拆分数组并将第一部分添加到末尾

2021年3月21日16:44:25 发表评论 612 次浏览

本文概述

有一个给定的数组, 并将其从指定位置拆分, 然后将数组add的第一部分移至末尾。

拆分数组并将第一部分添加到末尾

例子：

Input : arr[] = {12, 10, 5, 6, 52, 36}
            k = 2
Output : arr[] = {5, 6, 52, 36, 12, 10}
Explanation : Split from index 2 and first 
part {12, 10} add to the end .

Input : arr[] = {3, 1, 2}
           k = 1
Output : arr[] = {1, 2, 3}
Explanation : Split from index 1 and first
part add to the end.

推荐：请尝试以下方法{IDE}首先, 在继续解决方案之前。

简单的解决方案

我们一一旋转数组。

C ++

// CPP program to split array and move first
// part to end.
#include <bits/stdc++.h>
using namespace std;
  
void splitArr( int arr[], int n, int k)
{
     for ( int i = 0; i < k; i++) {
  
         // Rotate array by 1.
         int x = arr[0];
         for ( int j = 0; j < n - 1; ++j)
             arr[j] = arr[j + 1];
         arr[n - 1] = x;
     }
}
  
// Driver code
int main()
{
     int arr[] = { 12, 10, 5, 6, 52, 36 };
     int n = sizeof (arr) / sizeof (arr[0]);
     int position = 2;
  
     splitArr(arr, 6, position);
  
     for ( int i = 0; i < n; ++i)
         printf ( "%d " , arr[i]);
  
     return 0;
}

Java

// Java program to split array and move first
// part to end.
  
import java.util.*;
import java.lang.*;
class GFG {
     public static void splitArr( int arr[], int n, int k)
     {
         for ( int i = 0 ; i < k; i++) {
  
             // Rotate array by 1.
             int x = arr[ 0 ];
             for ( int j = 0 ; j < n - 1 ; ++j)
                 arr[j] = arr[j + 1 ];
             arr[n - 1 ] = x;
         }
     }
  
     // Driver code
     public static void main(String[] args)
     {
         int arr[] = { 12 , 10 , 5 , 6 , 52 , 36 };
         int n = arr.length;
         int position = 2 ;
  
         splitArr(arr, 6 , position);
  
         for ( int i = 0 ; i < n; ++i)
             System.out.print(arr[i] + " " );
     }
}
  
// Code Contributed by Mohit Gupta_OMG <(0_o)>

Python3

# Python program to split array and move first
# part to end.
  
def splitArr(arr, n, k): 
     for i in range ( 0 , k): 
         x = arr[ 0 ]
         for j in range ( 0 , n - 1 ):
             arr[j] = arr[j + 1 ]
          
         arr[n - 1 ] = x
          
  
# main
arr = [ 12 , 10 , 5 , 6 , 52 , 36 ]
n = len (arr)
position = 2
  
splitArr(arr, n, position)
  
for i in range ( 0 , n): 
     print (arr[i], end = ' ' )
  
# Code Contributed by Mohit Gupta_OMG <(0_o)>

C#

// C# program to split array 
// and move first part to end.
using System;
  
class GFG {
      
     // Function to spilt array and
     // move first part to end
     public static void splitArr( int [] arr, int n, int k)
     {
         for ( int i = 0; i < k; i++) 
         {
  
             // Rotate array by 1.
             int x = arr[0];
             for ( int j = 0; j < n - 1; ++j)
                 arr[j] = arr[j + 1];
             arr[n - 1] = x;
         }
     }
  
     // Driver code
     public static void Main()
     {
         int [] arr = {12, 10, 5, 6, 52, 36};
         int n = arr.Length;
         int position = 2;
         splitArr(arr, 6, position);
  
         for ( int i = 0; i < n; ++i)
             Console.Write(arr[i] + " " );
     }
}
  
// This code is contributed by Shrikant13.

的PHP

<?php
// PHP program to split array 
// and move first part to end.
  
function splitArr(& $arr , $n , $k )
{
     for ( $i = 0; $i < $k ; $i ++) 
     {
  
         // Rotate array by 1.
         $x = $arr [0];
         for ( $j = 0; $j < $n - 1; ++ $j )
             $arr [ $j ] = $arr [ $j + 1];
         $arr [ $n - 1] = $x ;
     }
}
  
// Driver code
$arr = array (12, 10, 5, 6, 52, 36);
$n = sizeof( $arr );
$position = 2;
  
splitArr( $arr , 6, $position );
  
for ( $i = 0; $i < $n ; ++ $i )
     echo $arr [ $i ]. " " ;
  
// This code is contributed
// by ChitraNayal
?>

输出如下：

5 6 52 36 12 10

上述解决方案的时间复杂度为O(nk)。

另一种方法：

另一种方法是制作一个具有两倍大小的临时数组, 然后将数组元素复制到新数组两次。然后, 以旋转为起始索引, 将元素从新数组复制到我们的数组, 直到数组的长度。

下面是上述方法的实现。

C ++

// CPP program to split array and move first
// part to end.
#include <bits/stdc++.h>
using namespace std;
  
// Function to spilt array and 
// move first part to end 
void splitArr( int arr[], int length, int rotation)
{
     int tmp[length * 2] = {0};
  
     for ( int i = 0; i < length; i++)
     {
         tmp[i] = arr[i];
         tmp[i + length] = arr[i];
     }
  
     for ( int i = rotation; i < rotation + length; i++)
     {
         arr[i - rotation] = tmp[i];
     }
}
  
// Driver code
int main()
{
     int arr[] = { 12, 10, 5, 6, 52, 36 };
     int n = sizeof (arr) / sizeof (arr[0]);
     int position = 2;
  
     splitArr(arr, n, position);
  
     for ( int i = 0; i < n; ++i)
         printf ( "%d " , arr[i]);
  
     return 0;
}
  
// This code is contributed by YashKhandelwal8

Java

// Java program to split array and move first
// part to end.
import java.util.*;
import java.lang.*;
class GFG {
      
     // Function to spilt array and
     // move first part to end
     public static void SplitAndAdd( int [] A, int length, int rotation){
          
         //make a temporary array with double the size 
         int [] tmp = new int [length* 2 ];
          
         // copy array element in to new array twice
         System.arraycopy(A, 0 , tmp, 0 , length);
         System.arraycopy(A, 0 , tmp, length, length);
         for ( int i=rotation;i<rotation+length;i++)
             A[i-rotation]=tmp[i];
     }
  
  
     // Driver code
     public static void main(String[] args)
     {
         int arr[] = { 12 , 10 , 5 , 6 , 52 , 36 };
         int n = arr.length;
         int position = 2 ;
  
         SplitAndAdd(arr, n, position);
  
         for ( int i = 0 ; i < n; ++i)
             System.out.print(arr[i] + " " );
     }
}

Python3

# Python3 program to split array and 
# move first part to end.
  
# Function to spilt array and 
# move first part to end
def SplitAndAdd(A, length, rotation):
  
     # make a temporary array with double 
     # the size and each index is initialized to 0 
     tmp = [ 0 for i in range (length * 2 )] 
  
     # copy array element in to new array twice
     for i in range (length):
         tmp[i] = A[i]
         tmp[i + length] = A[i]
  
     for i in range (rotation, rotation + length, 1 ): 
         A[i - rotation] = tmp[i]; 
       
# Driver code 
arr = [ 12 , 10 , 5 , 6 , 52 , 36 ] 
n = len (arr) 
position = 2
SplitAndAdd(arr, n, position); 
for i in range (n):
     print (arr[i], end = " " )
print () 
  
# This code is contributed by SOUMYA SEN

C#

// C# program to split array 
// and move first part to end. 
using System;
  
class GFG
{
  
     // Function to spilt array and 
     // move first part to end 
     public static void SplitAndAdd( int [] A, int length, int rotation)
     {
  
         // make a temporary array with double the size 
         int [] tmp = new int [length * 2];
  
         // copy array element in to new array twice 
         Array.Copy(A, 0, tmp, 0, length);
         Array.Copy(A, 0, tmp, length, length);
         for ( int i = rotation; i < rotation + length; i++)
         {
             A[i - rotation] = tmp[i];
         }
     }
  
     // Driver code 
     public static void Main( string [] args)
     {
         int [] arr = new int [] {12, 10, 5, 6, 52, 36};
         int n = arr.Length;
         int position = 2;
  
         SplitAndAdd(arr, n, position);
  
         for ( int i = 0; i < n; ++i)
         {
             Console.Write(arr[i] + " " );
         }
     }
}
  
// This code is contributed by kumar65

输出如下：

5 6 52 36 12 10

以下文章讨论了一种有效的O(n)解决方案：拆分数组并将第一部分添加到末尾|套装2

这个问题只不过是数组旋转问题, 我们可以在此处应用优化的O(n)数组旋转方法。

数组旋转程序

阵列旋转的块交换算法

阵列旋转的逆向算法

快速找到数组的多个左旋转|套装1

在O(n)时间和O(1)空间中打印数组的左旋转

如果发现任何不正确的地方, 或者想分享有关上述主题的更多信息, 请写评论。

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