递归程序以给定总和打印所有子集

给定一个数组和一个数字，请打印总和等于给定总和的所有子集。

例子：

Input :  arr[] =  {2, 5, 8, 4, 6, 11}, sum = 13
Output : 
5 8
2 11
2 5 6

Input : arr[] =  {1, 5, 8, 4, 6, 11}, sum = 9
Output :
5 4
1 8

此问题是检查是否存在具有给定总和的子集的扩展。我们递归地生成所有子集。我们跟踪当前子集的元素。如果当前子集中的元素总和等于给定总和，我们将打印该子集。

C++

// CPP program to print all subsets with given sum
#include 
using namespace std;
  
// The vector v stores current subset.
void printAllSubsetsRec(int arr[], int n, vector v,
                        int sum)
{
    // If remaining sum is 0, then print all
    // elements of current subset.
    if (sum == 0) {
        for (auto x : v)
            cout << x << " ";
        cout << endl;
        return;
    }
  
    // If no remaining elements,
    if (n == 0)
        return;
  
    // We consider two cases for every element.
    // a) We do not include last element.
    // b) We include last element in current subset.
    printAllSubsetsRec(arr, n - 1, v, sum);
    v.push_back(arr[n - 1]);
    printAllSubsetsRec(arr, n - 1, v, sum - arr[n - 1]);
}
  
// Wrapper over printAllSubsetsRec()
void printAllSubsets(int arr[], int n, int sum)
{
    vector v;
    printAllSubsetsRec(arr, n, v, sum);
}
  
// Driver code
int main()
{
    int arr[] = { 2, 5, 8, 4, 6, 11 };
    int sum = 13;
    int n = sizeof(arr) / sizeof(arr[0]);
    printAllSubsets(arr, n, sum);
    return 0;
}

Java

// Java program to print all subsets with given sum 
import java.util.*;
 class Solution
{
  
// The vector v stores current subset. 
static void printAllSubsetsRec(int arr[], int n, Vector v, 
                        int sum) 
{ 
    // If remaining sum is 0, then print all 
    // elements of current subset. 
    if (sum == 0) { 
        for (int i=0;i v1=new Vector(v);
    v1.add(arr[n - 1]); 
    printAllSubsetsRec(arr, n - 1, v1, sum - arr[n - 1]); 
} 
  
// Wrapper over printAllSubsetsRec() 
static void printAllSubsets(int arr[], int n, int sum) 
{ 
    Vector v= new Vector(); 
    printAllSubsetsRec(arr, n, v, sum); 
} 
  
// Driver code 
public static void main(String args[]) 
{ 
    int arr[] = { 2, 5, 8, 4, 6, 11 }; 
    int sum = 13; 
    int n = arr.length; 
    printAllSubsets(arr, n, sum); 
      
}
} 
//contributed by Arnab Kundu

Python3

# Python program to print all subsets with given sum
  
# The vector v stores current subset.
def printAllSubsetsRec(arr, n, v, sum) :
  
    # If remaining sum is 0, then print all
    # elements of current subset.
    if (sum == 0) :
        for value in v :
            print(value, end=" ")
        print()
        return
      
  
    # If no remaining elements,
    if (n == 0):
        return
  
    # We consider two cases for every element.
    # a) We do not include last element.
    # b) We include last element in current subset.
    printAllSubsetsRec(arr, n - 1, v, sum)
    v1 = [] + v
    v1.append(arr[n - 1])
    printAllSubsetsRec(arr, n - 1, v1, sum - arr[n - 1])
  
  
# Wrapper over printAllSubsetsRec()
def printAllSubsets(arr, n, sum):
  
    v = []
    printAllSubsetsRec(arr, n, v, sum)
  
  
# Driver code
  
arr = [ 2, 5, 8, 4, 6, 11 ]
sum = 13
n = len(arr)
printAllSubsets(arr, n, sum)
  
# This code is contributed by ihritik

C#

// C# program to print all subsets with given sum 
using System;
using System.Collections.Generic;
  
class GFG 
{ 
    // The vector v stores current subset. 
    static void printAllSubsetsRec(int []arr, int n, 
                                    List v, int sum) 
    { 
        // If remaining sum is 0, then print all 
        // elements of current subset. 
        if (sum == 0)
        { 
            for (int i = 0; i < v.Count; i++) 
                Console.Write( v[i]+ " "); 
            Console.WriteLine(); 
            return; 
        } 
  
        // If no remaining elements, 
        if (n == 0) 
            return; 
  
        // We consider two cases for every element. 
        // a) We do not include last element. 
        // b) We include last element in current subset. 
        printAllSubsetsRec(arr, n - 1, v, sum); 
        List v1 = new List(v); 
        v1.Add(arr[n - 1]); 
        printAllSubsetsRec(arr, n - 1, v1, sum - arr[n - 1]); 
    } 
  
    // Wrapper over printAllSubsetsRec() 
    static void printAllSubsets(int []arr, int n, int sum) 
    { 
        List v = new List(); 
        printAllSubsetsRec(arr, n, v, sum); 
    } 
  
    // Driver code 
    public static void Main() 
    { 
        int []arr = { 2, 5, 8, 4, 6, 11 }; 
        int sum = 13; 
        int n = arr.Length; 
        printAllSubsets(arr, n, sum); 
    } 
} 
  
// This code is contributed by Rajput-Ji

输出：

8 5 
6 5 2 
11 2

时间复杂度： O(2 ⁿ )

请参考下面的帖子，以获得基于动态编程的优化解决方案。使用动态编程打印给定总和的所有子集