以给定大小的组反转数组 |组3（单次遍历）

给定一个数组，反转由连续 k 个元素组成的每个子数组。

例子：

Input: arr = [1, 2, 3, 4, 5, 6, 7, 8, 9, 10], k = 3. 
Output: [3, 2, 1, 6, 5, 4, 9, 8, 7, 10]
Input: arr = [1, 2, 3, 4, 5, 6, 7], k = 5. 
Output: [5, 4, 3, 2, 1, 7, 6]

方法：

1. 我们将使用两个指针技术来解决这个问题。
2. 首先，我们将用值 k-1 ( d = k-1 ) 和一个值为 2 ( m = 2 ) 的变量 m 初始化我们的第一个指针 d。
3. 现在，我们将用我们的第二个指针 i 迭代数组并检查
   • 如果i < d ，则交换 (arr[i], arr[d]) 并将 d 减 1。否则，
   • 使d = k * m – 1 ， i = k * (m – 1) – 1和m = m + 1。

下面是上述方法的实现。

C++

// C++ program to reverse every sub-array
// formed by consecutive k elements
#include
using namespace std;
 
// Function to reverse every sub-array
// formed by consecutive k elements
void ReverseInGroup(int arr[], int n, int k)
{
    if(n < k)
    {
        k = n;
    }
 
    // Initialize variables
    int d = k - 1, m = 2;
    int i = 0;
         
    for(i = 0; i < n; i++)
    {
        
       if (i >= d)
       {
           // Update the variables
           d = k * (m);
           if(d >= n)
           {
               d = n;        
           }    
           i = k * (m - 1) - 1;
           m++;
       }
       else
       {
           int t = arr[i];
           arr[i] = arr[d];
           arr[d] = t;
       }
       d = d - 1;
    }
    return;
}
 
// Driver code
int main()
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int k = 3;
    int n = sizeof(arr) / sizeof(arr[0]);
 
    ReverseInGroup(arr, n, k);
 
    for(int i = 0; i < n; i++)
       cout << arr[i] << " ";
 
    return 0;
}
 
// This code is contributed by Code_Mech

C

// C program to reverse every sub-array
// formed by consecutive k elements
#include
 
// Function to reverse every sub-array
// formed by consecutive k elements
void ReverseInGroup(int arr[], int n, int k)
{
    if(n= d)
        {  
            // Update the variables
            d = k * (m);
            if(d>=n)
            {
                d = n;            
            }          
            i = k * (m - 1)-1;
            m++;
             
        }
        else
        {
            int t = arr[i];
            arr[i] = arr[d];
            arr[d] = t;
        }
 
     d = d - 1;     
        
    }
    return;
         
}
 
// Driver code
int main()
{
    int arr[] = {1, 2, 3, 4, 5, 6, 7};
    int k = 3;
   
    int n = sizeof(arr) / sizeof(arr[0]);
   
    ReverseInGroup(arr, n, k);
   
    for (int i = 0; i < n; i++)
        printf("%d ", 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 ReverseInGroup(int arr[],
                           int n, int k)
{
    if(n < k)
    {
        k = n;
    }
 
    // Initialize variables
    int d = k - 1, m = 2;
    int i = 0;
         
    for(i = 0; i < n; i++)
    {
       if (i >= d)
       {
            
           // Update the variables
           d = k * (m);
           if(d >= n)
           {
               d = n;            
           }        
           i = k * (m - 1) - 1;
           m++;
       }
       else
       {
           int t = arr[i];
           arr[i] = arr[d];
           arr[d] = t;
       }
       d = d - 1;    
    }
    return;
}
 
// Driver code
public static void main(String[] args)
{
    int arr[] = { 1, 2, 3, 4, 5, 6, 7 };
    int k = 3;
    int n = arr.length;
     
    ReverseInGroup(arr, n, k);
     
    for(int i = 0; i < n; i++)
       System.out.printf("%d ", arr[i]);
}
}
 
// This code is contributed by sapnasingh4991

Python3

# Python3 program to reverse
# every sub-array formed by
# consecutive k elements
 
# Function to reverse every
# sub-array formed by consecutive
# k elements
def ReverseInGroup(arr, n, k):
 
    if(n < k):
        k = n
 
    # Initialize variables
    d = k - 1
    m = 2
    i = 0
 
    while i < n:
        if (i >= d):
 
            # Update the
            # variables
            d = k * (m)
             
            if(d >= n):
                d = n
 
            i = k * (m - 1) - 1
            m += 1
 
        else:
            arr[i], arr[d] = (arr[d],
                              arr[i])
        d = d - 1
        i += 1
 
    return
 
# Driver code
if __name__ == "__main__":
 
    arr = [1, 2, 3,
           4, 5, 6, 7]
    k = 3
    n = len(arr)
    ReverseInGroup(arr, n, k)
 
    for i in range(n):
        print(arr[i],
              end = " ")
 
# This code is contributed by Chitranayal

C#

// 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
static void ReverseInGroup(int []arr,
                           int n, int k)
{
    if(n < k)
    {
        k = n;
    }
 
    // Initialize variables
    int d = k - 1, m = 2;
    int i = 0;
         
    for(i = 0; i < n; i++)
    {
        if (i >= d)
        {
                 
            // Update the variables
            d = k * (m);
            if(d >= n)
            {
                d = n;            
            }        
            i = k * (m - 1) - 1;
            m++;
        }
        else
        {
            int t = arr[i];
            arr[i] = arr[d];
            arr[d] = t;
        }
        d = d - 1;    
    }
    return;
}
 
// Driver code
public static void Main()
{
    int []arr = { 1, 2, 3, 4, 5, 6, 7 };
    int k = 3;
    int n = arr.Length;
     
    ReverseInGroup(arr, n, k);
     
    for(int i = 0; i < n; i++)
        Console.Write(arr[i] + " ");
}
}
 
// This code is contributed by Code_Mech

Javascript

// Javascript program to reverse every sub-array
// formed by consecutive k elements
 
 
// Function to reverse every sub-array
// formed by consecutive k elements
function ReverseInGroup(arr, n, k)
{
    if(n < k)
    {
        k = n;
    }
 
    // Initialize variables
    let d = k - 1, m = 2;
    let i = 0;
         
    for(i = 0; i < n; i++)
    {
         
        if (i >= d)
        {
            // Update the variables
            d = k * (m);
            if(d >= n)
            {
              d = n;       
            }   
            i = k * (m - 1) - 1;
            m++;
        }
        else
        {
            let t = arr[i];
            arr[i] = arr[d];
            arr[d] = t;
        }
        d = d - 1;
    }
    return;
}
 
// Driver code
 
    let arr = [ 1, 2, 3, 4, 5, 6, 7 ];
    let k = 3;
    let n = arr.length;
 
    ReverseInGroup(arr, n, k);
 
    for(let i = 0; i < n; i++)
    document.write(arr[i] + " ");
 
 
// This code is contributed by _saurabh_jaiswal

输出：

3 2 1 6 5 4 7

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