在不使用数组或循环的情况下打印{1,2,3，…n}的所有子集

给定自然数n ，打印集合的所有子集$\{1, 2, 3, ..., n\}$不使用任何数组或循环(仅允许使用递归)。

例子：

Input : n = 4
Output : { 1 2 3 4 }
         { 1 2 3 }
         { 1 2 4 }
         { 1 2 }
         { 1 3 4 }
         { 1 3 }
         { 1 4 }
         { 1 }
         { 2 3 4 }
         { 2 3 }
         { 2 4 }
         { 2 }
         { 3 4 }
         { 3 }
         { 4 }
         { }

Input : n = 2
Output : { 1 2 }
         { 1 }
         { 2 }
         { }

方法：

• 从…开始$num = 2^n - 1$最高为0。
• 考虑具有n位的num的二进制表示形式。
• 从代表1的最左位开始，第二位代表2，依此类推，直到代表n的第n位。
• 打印与该位对应的数字(如果已设置)。
• 对num的所有值执行上述步骤，直到它等于0。

让我们通过一个例子来理解上述方法：

考虑输入n = 4，从$num = 2^n - 1 = 15$ 。

依此类推…直到num = 0。

下面是上述方法的实现：

C++

// C++ code to print all subsets
// of {1, 2, 3, n} without using
// array or loop, just recursion.
#include 
using namespace std;
  
void subset(int, int, int);
  
// This recursive function calls subset
// function to print the subsets one by one.
// numBits --> number of bits needed to
// represent the number (simply input value n).
// num --> Initially equal to 2 ^ n - 1 and
// decreases by 1 every recursion until 0.
void printSubsets(int numOfBits, int num) 
{
    if (num >= 0)
    {
        cout << "{ ";
  
        // Print the subset corresponding to
        // binary representation of num.
        subset(numOfBits - 1, num, numOfBits);
        cout << "}" << endl;
  
        // Call the function recursively to
        // print the next subset.
        printSubsets(numOfBits, num - 1);
    }
    else
        return;
}
  
// This function recursively prints the
// subset corresponding to the binary
// representation of num.
// nthBit --> nth bit from right side
// starting from n and decreases until 0
void subset(int nthBit, int num, int numOfBits)
{
    if (nthBit >= 0) 
    {
        // Print number in given subset only
        // if the bit corresponding to it
        // is set in num.
        if (num & (1 << nthBit)) 
        {
            cout << numOfBits - nthBit << " ";
        }
  
        // Check for the next bit
        subset(nthBit - 1, num, numOfBits);
    } 
    else
        return;
}
  
// Driver Code
int main()
{
    int n = 4;
    printSubsets(n, pow(2, n) - 1);
}
  
// This code is contributed by
// sanjeev2552

Java

// Java code to print all subsets 
// of {1, 2, 3, n} without using
// array or loop, just recursion.
class GfG 
{
  
    // This recursive function calls subset
    // function to print the subsets one by one. 
    // numBits --> number of bits needed to 
    // represent the number (simply input value n).
    // num --> Initially equal to 2 ^ n - 1 and 
    // decreases by 1 every recursion until 0.
    static void printSubSets(int numOfBits, int num) 
    {
        if (num >= 0) 
        {
            System.out.print("{ ");
              
            // Print the subset corresponding to 
            // binary representation of num.
            subset(numOfBits - 1, num, numOfBits);
            System.out.println("}");
              
            // Call the function recursively to 
            // print the next subset.
            printSubSets(numOfBits, num - 1);
  
        } else
            return;
    }
  
    // This function recursively prints the 
    // subset corresponding to the binary 
    // representation of num.
    // nthBit --> nth bit from right side 
    // starting from n and decreases until 0.
    static void subset(int nthBit, int num, int numOfBits) 
    {
        if (nthBit >= 0) 
        {
            // Print number in given subset only
            // if the bit corresponding to it 
            // is set in num.
            if ((num & (1 << nthBit)) != 0)
            {
                System.out.print(numOfBits - nthBit + " ");
  
            }
              
            // Check for the next bit 
            subset(nthBit - 1, num, numOfBits);
        } else
            return;
    }
      
    // Driver code
    public static void main(String[] args) 
    {
        int n = 4;
        printSubSets(n, (int) (Math.pow(2, n)) -1);
    }
}
  
// This code is contributed by laststringx

Python3

# Python3 code to print all subsets 
# of {1, 2, 3, …n} without using
# array or loop, just recursion.
  
# This recursive function calls subset
# function to print the subsets one by one. 
# numBits --> number of bits needed to 
# represent the number (simply input value n).
# num --> Initially equal to 2 ^ n - 1 and 
# decreases by 1 every recursion until 0.
def printSubsets(numOfBits, num):
      
    if num >= 0:
        print("{", end = " ")
  
        # Print the subset corresponding to 
        # binary representation of num.
        subset(numOfBits-1, num, numOfBits)
        print("}")
  
        # Call the function recursively to 
        # print the next subset.
        printSubsets(numOfBits, num-1)
          
    else:
        return
  
# This function recursively prints the 
# subset corresponding to the binary 
# representation of num.
# nthBit --> nth bit from right side 
# starting from n and decreases until 0.
def subset(nthBit, num, numOfBits):
      
    if nthBit >= 0:
          
        # Print number in given subset only
        # if the bit corresponding to it 
        # is set in num.
        if num & (1 << nthBit) != 0:
            print(numOfBits - nthBit, end = " ")
          
        # Check for the next bit 
        subset(nthBit-1, num, numOfBits)
          
    else:
        return
  
# Driver Code    
n = 4
printSubsets(n, 2**n - 1)

C#

// C# code to print all subsets 
// of {1, 2, 3, n} without using 
// array or loop, just recursion.
using System;
  
class GfG 
{ 
  
    // This recursive function calls subset 
    // function to print the subsets one by one. 
    // numBits --> number of bits needed to 
    // represent the number (simply input value n). 
    // num --> Initially equal to 2 ^ n - 1 and 
    // decreases by 1 every recursion until 0. 
    static void printSubSets(int numOfBits, int num) 
    { 
        if (num >= 0) 
        { 
            Console.Write("{ "); 
              
            // Print the subset corresponding to 
            // binary representation of num. 
            subset(numOfBits - 1, num, numOfBits); 
            Console.WriteLine("}"); 
              
            // Call the function recursively to 
            // print the next subset. 
            printSubSets(numOfBits, num - 1); 
  
        } else
            return; 
    } 
  
    // This function recursively prints the 
    // subset corresponding to the binary 
    // representation of num. 
    // nthBit --> nth bit from right side 
    // starting from n and decreases until 0. 
    static void subset(int nthBit, int num, int numOfBits) 
    { 
        if (nthBit >= 0) 
        { 
            // Print number in given subset only 
            // if the bit corresponding to it 
            // is set in num. 
            if ((num & (1 << nthBit)) != 0) 
            { 
                Console.Write(numOfBits - nthBit + " "); 
  
            } 
              
            // Check for the next bit 
            subset(nthBit - 1, num, numOfBits); 
        } else
            return; 
    } 
      
    // Driver codeM
    public static void Main(String[] args) 
    { 
        int n = 4; 
        printSubSets(n, (int) (Math.Pow(2, n)) -1); 
    } 
}
  
// This code is contributed by Srathore

输出：

{ 1 2 3 4 }
{ 1 2 3 }
{ 1 2 4 }
{ 1 2 }
{ 1 3 4 }
{ 1 3 }
{ 1 4 }
{ 1 }
{ 2 3 4 }
{ 2 3 }
{ 2 4 }
{ 2 }
{ 3 4 }
{ 3 }
{ 4 }
{ }

时间复杂度： $O(n*2^n)$