算法（拆分数组并将第一部分添加到末尾）

本文概述

建议：在继续解决方案之前, 请先在{IDE}上尝试使用你的方法。
C ++
Java
Python3
C#
的PHP
C ++
Java
Python3
C#

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。

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例子：

Input : arr[] = {12, 10, 5, 6, 52, 36}k = 2Output : 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 = 1Output : arr[] = {1, 2, 3}Explanation : Split from index 1 and firstpart 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 = 2splitArr(arr, n, position)for i in range ( 0 , n): print (arr[i], end = ' ' )# Code Contributed by Mohit Gupta_OMG < (0_o)>

// 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# 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

【算法（拆分数组并将第一部分添加到末尾）】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。