使用递归生成子数组

给定一个数组，使用递归生成给定数组的所有可能的子数组。

例子：

Input : [1, 2, 3]
Output : [1], [1, 2], [2], [1, 2, 3], [2, 3], [3]

Input : [1, 2]
Output : [1], [1, 2], [2]

我们已经讨论了生成所有子数组的迭代程序。在这篇文章中，讨论了递归。

方法：我们使用两个指针start和end来维护数组的起点和终点，并遵循以下步骤：

• 如果我们已经到达数组的末尾，则停止
• 如果开始大于结束，则增加结束索引
• 从索引开始到结束打印子数组并增加起始索引

下面是上述方法的实现。

C++

// C++ code to print all possible subarrays  
// for given array using recursion 
  
#include 
# include 
using namespace std;
  
// Recursive function to print all possible subarrays  
// for given array 
void printSubArrays(vector arr, int start, int end)
{      
    // Stop if we have reached the end of the array     
    if (end == arr.size()) 
        return;
        
    // Increment the end point and start from 0 
    else if (start > end) 
        printSubArrays(arr, 0, end + 1);
            
    // Print the subarray and increment the starting point 
    else
    {
        cout << "[";
        for (int i = start; i < end; i++){
            cout << arr[i] << ", ";
        }
          
        cout << arr[end] << "]" << endl;
        printSubArrays(arr, start + 1, end);
    }
      
    return;
}
  
int main()
{
   vector arr = {1, 2, 3};
   printSubArrays(arr, 0, 0);
   return 0;
}

Java

// Java code to print all possible subarrays 
// for given array using recursion 
  
class solution
{
  
// Recursive function to print all possible subarrays 
// for given array 
static void printSubArrays(int []arr, int start, int end)
{     
    // Stop if we have reached the end of the array     
    if (end == arr.length) 
        return;
      
    // Increment the end point and start from 0 
    else if (start > end) 
        printSubArrays(arr, 0, end + 1);
          
    // Print the subarray and increment the starting point 
    else
    {
        System.out.print("[");
        for (int i = start; i < end; i++){
            System.out.print(arr[i]+", ");
        }
          
        System.out.println(arr[end]+"]");
        printSubArrays(arr, start + 1, end);
    }
      
    return;
}
  
public static void main(String args[])
{
int []arr = {1, 2, 3};
printSubArrays(arr, 0, 0);
  
}
}

Python3

# Python3 code to print all possible subarrays 
# for given array using recursion
  
# Recursive function to print all possible subarrays 
# for given array
def printSubArrays(arr, start, end):
      
    # Stop if we have reached the end of the array    
    if end == len(arr):
        return
      
    # Increment the end point and start from 0
    elif start > end:
        return printSubArrays(arr, 0, end + 1)
          
    # Print the subarray and increment the starting
    # point
    else:
        print(arr[start:end + 1])
        return printSubArrays(arr, start + 1, end)
          
# Driver code
arr = [1, 2, 3]
printSubArrays(arr, 0, 0)

C#

// C# code to print all possible subarrays 
// for given array using recursion 
using System;
  
class GFG 
{ 
  
    // Recursive function to print all  
    // possible subarrays for given array 
    static void printSubArrays(int []arr, 
                        int start, int end) 
    { 
        // Stop if we have reached 
        // the end of the array 
        if (end == arr.Length) 
            return; 
  
        // Increment the end point 
        // and start from 0 
        else if (start > end) 
            printSubArrays(arr, 0, end + 1); 
  
        // Print the subarray and 
        // increment the starting point 
        else
        { 
            Console.Write("["); 
            for (int i = start; i < end; i++)
            { 
                Console.Write(arr[i]+", "); 
            } 
  
            Console.WriteLine(arr[end]+"]"); 
            printSubArrays(arr, start + 1, end); 
        } 
        return; 
    } 
  
    // Driver code
    public static void Main(String []args) 
    { 
        int []arr = {1, 2, 3}; 
        printSubArrays(arr, 0, 0); 
    } 
} 
  
// This code is contributed by Rajput-Ji

PHP

 $end)
        return printSubArrays($arr, 0, 
                              $end + 1);
          
    // Print the subarray and increment 
    // the starting point
    else
    {
    echo "[";
    for($i = $start; $i < $end + 1; $i++)
    {
        echo $arr[$i];
        if($i != $end)
        echo ", ";
    }
    echo "]\n";
        return printSubArrays($arr, $start + 1, 
                                    $end);
    }
} 
  
// Driver code
$arr = array(1, 2, 3);
printSubArrays($arr, 0, 0);
  
// This code is contributed by mits
?>

输出：

[1]
[1, 2]
[2]
[1, 2, 3]
[2, 3]
[3]

时间复杂度： $O(n^2)$