查找数组中长度为k的所有子集

Given a set {1,2,3,4,5...n} of n elements, we need to find all subsets of length k .

给定一组{ 1、2、3、4、5……n}对于n个元素，我们需要找到所有长度为k的子集。

For example, if n = 4 and k = 2, the output would be {1, 2}, {1, 3}, {1, 4}, {2, 3}, {2, 4}, {3, 4}.

例如，如果n = 4和k = 2，输出将是{1,2}，{1,3}，{1,4}，{2,3}，{2,4}，{3,4}。

I am not even able to figure out how to start. We don't have to use the inbuilt library functions like next_permutation etc.

我甚至不知道怎么开始。我们不需要使用内建库函数比如next_permutation等等。

Need the algorithm and implementation in either C/C++ or Java.

需要C/ c++或Java中的算法和实现。

11 个解决方案

#1

Recursion is your friend for this task.

递归是这个任务的朋友。

For each element - "guess" if it is in the current subset, and recursively invoke with the guess and a smaller superset you can select from. Doing so for both the "yes" and "no" guesses - will result in all possible subsets.
Restraining yourself to a certain length can be easily done in a stop clause.

对于每个元素——如果它位于当前子集，则“guess”，并递归地使用guess和一个更小的超集进行调用。对于“是”和“不”的猜测，这样做将导致所有可能的子集。在stop子句中，把自己限制在一定的长度是很容易做到的。

Java code:

Java代码:

private static void getSubsets(List<Integer> superSet, int k, int idx, Set<Integer> current,List<Set<Integer>> solution) {
    //successful stop clause
    if (current.size() == k) {
        solution.add(new HashSet<>(current));
        return;
    }
    //unseccessful stop clause
    if (idx == superSet.size()) return;
    Integer x = superSet.get(idx);
    current.add(x);
    //"guess" x is in the subset
    getSubsets(superSet, k, idx+1, current, solution);
    current.remove(x);
    //"guess" x is not in the subset
    getSubsets(superSet, k, idx+1, current, solution);
}

public static List<Set<Integer>> getSubsets(List<Integer> superSet, int k) {
    List<Set<Integer>> res = new ArrayList<>();
    getSubsets(superSet, k, 0, new HashSet<Integer>(), res);
    return res;
}

Invoking with:

调用:

List<Integer> superSet = new ArrayList<>();
superSet.add(1);
superSet.add(2);
superSet.add(3);
superSet.add(4);
System.out.println(getSubsets(superSet,2));

Will yield:

将收益率:

[[1, 2], [1, 3], [1, 4], [2, 3], [2, 4], [3, 4]]

#2

Use a bit vector representation of the set, and use an algorithm similar to what std::next_permutation does on 0000.1111 (n-k zeroes, k ones). Each permutation corresponds to a subset of size k.

使用集合的位向量表示，并使用类似于std::next_permutation在0000.1111 (n-k个0,k个1)上所做的算法。每个排列对应于大小为k的子集。

#3

Check out my solution

看看我的解决方案

import java.util.ArrayList;
import java.util.HashSet;
import java.util.Set;


 public class Subset_K {
public static void main(String[]args)
{
    Set<String> x;
    int n=4;
    int k=2;
    int arr[]={1,2,3,4};
    StringBuilder sb=new StringBuilder();
    for(int i=1;i<=(n-k);i++)
        sb.append("0");
    for(int i=1;i<=k;i++)
        sb.append("1");
    String bin=sb.toString();
    x=generatePerm(bin);
    Set<ArrayList <Integer>> outer=new HashSet<ArrayList <Integer>>();
    for(String s:x){
        int dec=Integer.parseInt(s,2);
        ArrayList<Integer> inner=new ArrayList<Integer>();
        for(int j=0;j<n;j++){
            if((dec&(1<<j))>0)
                inner.add(arr[j]);
        }
        outer.add(inner);
    }
    for(ArrayList<?> z:outer){
        System.out.println(z);
    }
}

    public static Set<String> generatePerm(String input)
{
    Set<String> set = new HashSet<String>();
    if (input == "")
        return set;

    Character a = input.charAt(0);

    if (input.length() > 1)
    {
        input = input.substring(1);

        Set<String> permSet = generatePerm(input);

        for (String x : permSet)
        {
            for (int i = 0; i <= x.length(); i++)
            {
                set.add(x.substring(0, i) + a + x.substring(i));
            }
        }
    }
    else
    {
        set.add(a + "");
    }
    return set;
}
}

I am working on a 4 element set for test purpose and using k=2. What I try to do is initially generate a binary string where k bits are set and n-k bits are not set. Now using this string I find all the possible permutations of this string. And then using these permutations I output the respective element in the set. Would be great if someone could tell me about the complexity of this problem.

我正在为测试目的而设计一个4元素集，并使用k=2。我首先要做的是生成一个二进制字符串，其中k位被设置，n-k位未被设置。然后使用这些排列，我输出集合中相应的元素，如果有人能告诉我这个问题的复杂性那就太好了。

#4

This is python. Sorry for the spanish ;)

这是python。对不起，西班牙语;

from pprint import pprint
conjunto = [1,2,3,4, 5,6,7,8,9,10]
k = 3
lista = []
iteraciones = [0]
def subconjuntos(l, k):
    if k == len(l):
        if not l in lista:
            lista.append(l)
        return
    for i in l:
        aux = l[:]
        aux.remove(i)
        result = subconjuntos(aux, k)
        iteraciones[0] += 1
        if not result in lista and result:
            lista.append( result)

subconjuntos(conjunto, k)
print (lista)
print ('cant iteraciones: ' + str(iteraciones[0]))

#5

   #include<iostream>
   #include<cstdio>
   #include<vector>
   using namespace std;
   vector<int> v;
   vector<vector<int> > result;

  void subset(int arr[],int k,int n,int idx){
  if(idx==n)
 return;

if(k==1){
    for(int i=idx;i<n;i++)
     {
        v.push_back(arr[i]);
        result.push_back(v);
        v.pop_back();
     }
}

 for(int j=idx;j<n;j++) {
  v.push_back(arr[j]);
  subset(arr,k-1,n,j+1);
  v.pop_back();
  }
 }

int main(){
int arr[] = {1,2,3,4,5,6,7};
int k = 4;
int n =sizeof(arr)/sizeof(arr[0]);
subset(arr,k,n,0);

for(int i = 0;i<result.size();i++)
 { 
  for(int j = 0;j<result[i].size();j++)
   {
     cout << result[i][j] << " ";
   }
   cout << endl;
 }
}

#6

Please check my solution:-

请检查我的解决方案:-

private static void printPermutations(List<Integer> list, int subSetSize) {
    List<Integer> prefixList = new ArrayList<Integer>();
    printPermutations(prefixList, list, subSetSize);
}

private static void printPermutations(List<Integer> prefixList, List<Integer> list, int subSetSize) {
    if (prefixList.size() == subSetSize) {
        System.out.println(prefixList);
    } else {
        for (int i = 0; i < list.size(); i++) {
            Integer removed = list.remove(i);
            prefixList.add(removed);
            printPermutations(prefixList, list, subSetSize);
            prefixList.remove(removed);
            list.add(i, removed);
        }
    }
}

This is similar to String permutations:-

这类似于字符串排列:-

private static void printPermutations(String str) {
    printAllPermutations("", str);
}

private static void printAllPermutations(String prefix, String restOfTheString) {
    int len = restOfTheString.length();
    System.out.println(prefix);
    for (int i = 0; i < len; i++) {
        printAllPermutations(prefix + restOfTheString.charAt(i), restOfTheString.substring(0, i) + restOfTheString.substring(i + 1, len));
    }
}

#7

This is an implemation in F#:

这是f#的一个实现:

// allSubsets: int -> int -> Set<Set<int>>
let rec allSubsets n k =
    match n, k with
    | _, 0 -> Set.empty.Add(Set.empty)
    | 0, _ -> Set.empty
    | n, k -> Set.union (Set.map (fun s -> Set.add n s) (allSubsets (n-1) (k-1)))
                        (allSubsets (n-1) k)

You can try it in the F# REPL:

你可以在f# REPL中试试:

> allSubsets 3 2;;

val it : Set<Set<int>> = set [set [1; 2]; set [1; 3]; set [2; 3]]

> allSubsets 4 2;;

val it : Set<Set<int>> = set [set [1; 2]; set [1; 3]; set [1; 4]; set [2; 3]; set [2; 4]; set [3; 4]]

This Java class implements the same algorithm:

这个Java类实现了相同的算法:

import java.util.HashSet;
import java.util.Set;

public class AllSubsets {

    public static Set<Set<Integer>> allSubsets(int setSize, int subsetSize) {
        if (subsetSize == 0) {
            HashSet<Set<Integer>> result = new HashSet<>();
            result.add(new HashSet<>());
            return result;
        }
        if (setSize == 0) {
            return new HashSet<>();
        }
        Set<Set<Integer>> sets1 = allSubsets((setSize - 1), (subsetSize - 1));
        for (Set<Integer> set : sets1) {
            set.add(setSize);
        }
        Set<Set<Integer>> sets2 = allSubsets((setSize - 1), subsetSize);
        sets1.addAll(sets2);
        return sets1;
    }
}

If you do not like F# or Java then visit this website. It lists solutions to your particular problem in various programming languages:

如果你不喜欢f#或Java，请访问这个网站。它列出了用各种编程语言解决你的问题的方法:

http://rosettacode.org/wiki/Combinations

http://rosettacode.org/wiki/Combinations

#8

JavaScript implementation:

JavaScript实现:

var subsetArray = (function() {
  return {
    getResult: getResult
  }

  function getResult(array, n) {

    function isBigEnough(value) {
      return value.length === n;
    }

    var ps = [
      []
    ];
    for (var i = 0; i < array.length; i++) {
      for (var j = 0, len = ps.length; j < len; j++) {
        ps.push(ps[j].concat(array[i]));
      }
    }
    return ps.filter(isBigEnough);
  }
})();



 var arr = [1, 2, 3, 4,5,6,7,8,9];
 console.log(subsetArray.getResult(arr,2));

#9

Here is an iterative version in python. Essence of it is increment_counters() function which returns all possible combinations. We know it needs to be called C(n,r) times.

这是python中的一个迭代版本。它的本质是increment_counter()函数，它返回所有可能的组合。我们知道它需要被称为C(n,r)乘以。

def nchooser(n,r):
    """Calculate the n choose r manual way"""
    import math
    f = math.factorial
    return f(n) / f(n-r) / f(r)

def increment_counters(rc,r,n):
    """This is the essense of the algorithm. It generates all possible indexes.
    Ex: for n = 4, r = 2, rc will have values (0,1),(0,2),(0,3),(1,2),(1,3),(2,3).
    You may have better understanding if you print all possible 35 values for
    n = 7, r = 3."""

    rc[r-1] += 1     # first increment the least significant counter
    if rc[r-1] < n:  # if it does not overflow, return
        return

    # overflow at the last counter may cause some of previous counters to overflow
    # find where it stops (ex: in n=7,r=3 case, 1,2,3 will follow 0,5,6)
    for i in range(r-2,-1,-1): # from r-2 to 0 inclusive
        if rc[i] < i+n-r:
            break
    # we found that rc[i] will not overflow. So, increment it and reset the
    # counters right to it. 
    rc[i] += 1
    for j in range(i+1,r):
        rc[j] = rc[j-1] + 1

def combinations(lst, r):
    """Return all different sub-lists of size r"""
    n = len(lst)
    rc = [ i for i in range(r) ]  # initialize counters
    res = []
    for i in range(nchooser(n,r)): # increment the counters max possible times 
        res.append(tuple(map(lambda k: lst[k],rc)))
        increment_counters(rc,r,n)

    return res

#10

Here is a Java version of what I think Simple is talking about, using a binary representation of all sets in the power set. It's similar to how Abhiroop Sarkar did it, but I think a boolean array makes more sense than a string when you are just representing binary values.

这是一个Java版本的我认为简单的讨论,使用二进制表示集的幂集。它是类似于Abhiroop Sarkar做到了,但我认为一个布尔比一个字符串数组更有意义,当你只是代表二进制值。

private ArrayList<ArrayList<Object>> getSubsets(int m, Object[] objects){
    // m = size of subset, objects = superset of objects
    ArrayList<ArrayList<Object>> subsets = new ArrayList<>();
    ArrayList<Integer> pot = new ArrayList<>();
    int n = objects.length;
    int p = 1;
    if(m==0)
        return subsets;
    for(int i=0; i<=n; i++){
        pot.add(p);
        p*=2;
    }
    for(int i=1; i<p; i++){
        boolean[] binArray = new boolean[n];
        Arrays.fill(binArray, false);
        int y = i;
        int sum = 0;
        for(int j = n-1; j>=0; j--){
            int currentPot = pot.get(j);
            if(y >= currentPot){
                binArray[j] = true;
                y -= currentPot;
                sum++;
            }
            if(y<=0)
                break;
        }
        if(sum==m){
            ArrayList<Object> subsubset = new ArrayList<>();
            for(int j=0; j < n; j++){
                if(binArray[j]){
                    subsubset.add(objects[j]);
                }
            }
            subsets.add(subsubset);
        }
    }

    return subsets;
}

#11

If you are looking for Iterator pattern answer then here you go.

如果您正在寻找迭代器模式的答案，那么这里就是。

public static <T> Iterable<List<T>> getList(final Iterable<? extends T> list) {

    List<List<T>> listOfList = new ArrayList<>();

    for (T t: list)
        listOfList.add(Collections.singletonList(t));

    return listOfList;
}
public static <T> Iterable<List<T>> getIterable(final Iterable<? extends T> list, final int size) {

    final List<T> vals = new ArrayList<>();
    int numElements = 0;
    for (T t : list) {
        vals.add(t);
        numElements++;
    }

    if (size == 1) {
        return getList(vals);
    }
    if (size == numElements) {
        return Collections.singletonList(vals);
    }

    return new Iterable<List<T>>() {

        @Override
        public Iterator<List<T>> iterator() {
            return new Iterator<List<T>>() {

                int currPos = 0;                    
                Iterator<List<T>> nextIterator = getIterable(
                    vals.subList(this.currPos + 1, vals.size()), size - 1).iterator();

                @Override
                public boolean hasNext() {
                    if ((this.currPos < vals.size()-2) && (this.currPos+size < vals.size()))
                        return true;
                    return false;
                }

                @Override
                public List<T> next() {
                    if (!nextIterator.hasNext()) {
                        this.currPos++;
                        nextIterator = getIterable(vals.subList(this.currPos+1, vals.size()), size-1).iterator();
                    }
                    final List<T> ret = new ArrayList<>(nextIterator.next());
                    ret.add(0, vals.get(this.currPos));
                    return ret;
                }
            };
        }
    };
}