算法设计（如何反转给定大小的组中的数组（））

本文概述

建议：在继续解决方案之前, 请先在"实践"上解决它。
C ++
Java
Python3
C#
的PHP

给定一个数组, 反转由连续的k个元素形成的每个子数组。
例子：

输入：arr = [1、2、3、4、5、6、7、8、9] k = 3输出：[3、2、1、6、5、4、9、8、7]输入：arr = [1、2、3、4、5、6、7、8] k = 5输出：[5、4、3、2、1、8、7、6]输入：arr = [1、2、3 , 4, 5, 6] k = 1输出：[1, 2, 3, 4, 5, 6]输入：arr = [1, 2, 3, 4, 5, 6, 7, 8] k = 10输出：[8、7、6、5、4、3、2、1]

推荐：请在"实践首先, 在继续解决方案之前。方法：考虑大小的每个子数组?从数组的开头开始并将其反转。我们需要处理一些特殊情况。如果k不是n的倍数, 其中n是数组的大小, 那么对于最后一组, 我们剩下的元素少于k个, 我们需要反转所有剩余的元素。如果k = 1, 数组应保持不变。如果k> = n, 则反转数组中存在的所有元素。
下图是上述方法的模拟：

文章图片
下面是上述方法的实现：
C ++

// C++ program to reverse every sub-array formed by // consecutive k elements #include < iostream> using namespace std; // Function to reverse every sub-array formed by // consecutive k elements void reverse( int arr[], int n, int k) { for ( int i = 0; i < n; i += k) { int left = i; // to handle case when k is not multiple of n int right = min(i + k - 1, n - 1); // reverse the sub-array [left, right] while (left < right) swap(arr[left++], arr[right--]); } }// Driver code int main() { int arr[] = {1, 2, 3, 4, 5, 6, 7, 8}; int k = 3; int n = sizeof (arr) / sizeof (arr[0]); reverse(arr, n, k); for ( int i = 0; i < n; i++) cout < < arr[i] < < " " ; return 0; }

Java

// Java program to reverse every sub-array formed by // consecutive k elements class GFG {// Function to reverse every sub-array formed by // consecutive k elements static void reverse( int arr[], int n, int k) { for ( int i = 0 ; i < n; i += k) { int left = i; // to handle case when k is not multiple // of n int right = Math.min(i + k - 1 , n - 1 ); int temp; // reverse the sub-array [left, right] while (left < right) { temp=arr[left]; arr[left]=arr[right]; arr[right]=temp; left+= 1 ; right-= 1 ; } } }// Driver method public static void main(String[] args) {int arr[] = { 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 }; int k = 3 ; int n = arr.length; reverse(arr, n, k); for ( int i = 0 ; i < n; i++) System.out.print(arr[i] + " " ); } }// This code is contributed by Anant Agarwal.

Python3

# Python 3 program to reverse every # sub-array formed by consecutive k # elements# Function to reverse every sub-array # formed by consecutive k elements def reverse(arr, n, k): i = 0while (i< n):left = i # To handle case when k is not # multiple of n right = min (i + k - 1 , n - 1 ) # Reverse the sub-array [left, right] while (left < right):arr[left], arr[right] = arr[right], arr[left] left + = 1 ; right - = 1 i + = k# Driver code arr = [ 1 , 2 , 3 , 4 , 5 , 6 , 7 , 8 ] k = 3 n = len (arr) reverse(arr, n, k)for i in range ( 0 , n): print (arr[i], end = " " )# This code is contributed by Smitha Dinesh Semwal

// C# program to reverse every sub-array // formed by consecutive k elements using System; class GFG {// Function to reverse every sub-array // formed by consecutive k elements public static void reverse( int [] arr, int n, int k) { for ( int i = 0; i < n; i += k) { int left = i; // to handle case when k is // not multiple of n int right = Math.Min(i + k - 1, n - 1); int temp; // reverse the sub-array [left, right] while (left < right) { temp = arr[left]; arr[left] = arr[right]; arr[right] = temp; left += 1; right -= 1; } } }// Driver Code public static void Main( string [] args) { int [] arr = new int [] {1, 2, 3, 4, 5, 6, 7, 8}; int k = 3; int n = arr.Length; reverse(arr, n, k); for ( int i = 0; i < n; i++) { Console.Write(arr[i] + " " ); } } }// This code is contributed // by Shrikant13

的PHP

< ?php // PHP program to reverse every sub-array // formed by consecutive k elements// Function to reverse every sub-array // formed by consecutive k elements function reverse( $arr , $n , $k ) { for ( $i = 0; $i < $n ; $i += $k ) { $left = $i ; // to handle case when k is not // multiple of n $right = min( $i + $k - 1, $n - 1); $temp ; // reverse the sub-array [left, right] while ( $left < $right ) { $temp = $arr [ $left ]; $arr [ $left ] = $arr [ $right ]; $arr [ $right ] = $temp ; $left += 1; $right -= 1; } } return $arr ; }// Driver Code $arr = array (1, 2, 3, 4, 5, 6, 7, 8); $k = 3; $n = sizeof( $arr ); $arr1 = reverse( $arr , $n , $k ); for ( $i = 0; $i < $n ; $i ++) echo $arr1 [ $i ] . " " ; // This code is contributed // by Akanksha Rai ?>

输出如下：