打印所有可能的方式将数组拆分成 K 个子集
原文:https://www . geeksforgeeks . org/print-所有可能的方法-将数组拆分为 k 个子集/
给定一个大小为 N 的数组arr【】和一个整数 K ，任务是打印所有可能的方式将给定数组拆分为 K 子集。
示例:
输入: arr[] = { 1，2，3 }，K = 2 输出: { {{ 1，2 }，{ 3 }}，{{ 1，3 }，{ 2 }}，{{ 1 }，{ 2，3 }}。
输入: arr[] = { 1，2，3，4 }，K = 2 输出:，{ 4 }}，{{ 1，2，4 }，{ 3 }，{{ 1，2 }，{ 3，4 }}，{{ 1，3，4 }，{ 2 }，{{ 1，3 }，{ 2，4 }}，{ { 1，4 }，{ 2，3 }}，{{ 1 }，{ 2 }
方法:使用回溯生成并打印所有子集可以解决问题。按照以下步骤解决问题:
- 遍历数组并使用以下递归关系将元素插入任意一个 K 子集:
分区 ub(i，k，n)
{
for(j = 0；j < K；j++){ 0
sub[j]。推回(arr[i])
分区 ub(i + 1，k，n)
sub[j]。pop_back()
}
}
- 如果 K 等于 0 或 K > N ，则不能生成子集。
- 如果插入到 K 子集中的数组元素个数等于 N ，则打印该子集中的元素。
C++
// C++ program for the above approach
#include <bits/stdc++.h>
using namespace std;
// arr: Store input array
// i: Stores current index of arr
// N: Stores length of arr
// K: Stores count of subsets
// nos: Stores count of feasible subsets formed
// v: Store K subsets of the given array
// Utility function to find all possible
// ways to split array into K subsets
void PartitionSub(int arr[], int i,
int N, int K, int nos,
vector<vector<int> >& v)
{
// If count of elements in K subsets
// are greater than or equal to N
if (i >= N) {
// If count of subsets
// formed is equal to K
if (nos == K) {
// Print K subsets by splitting
// array into K subsets
for (int x = 0; x < v.size(); x++) {
cout << "{ ";
// Print current subset
for (int y = 0; y < v[x].size(); y++) {
cout << v[x][y];
// If current element is the last
// element of the subset
if (y == v[x].size() - 1) {
cout << " ";
}
// Otherwise
else {
cout << ", ";
}
}
if (x == v.size() - 1) {
cout << "}";
}
else {
cout << "}, ";
}
}
cout << endl;
}
return;
}
for (int j = 0; j < K; j++) {
// If any subset is occupied,
// then push the element
// in that first
if (v[j].size() > 0) {
v[j].push_back(arr[i]);
// Recursively do the same
// for remaining elements
PartitionSub(arr, i + 1, N, K, nos, v);
// Backtrack
v[j].pop_back();
}
// Otherwise, push it in an empty
// subset and increase the
// subset count by 1
else {
v[j].push_back(arr[i]);
PartitionSub(arr, i + 1, N, K, nos + 1, v);
v[j].pop_back();
// Break to avoid the case of going in
// other empty subsets, if available,
// and forming the same combination
break;
}
}
}
// Function to to find all possible ways to
// split array into K subsets
void partKSubsets(int arr[], int N, int K)
{
// Stores K subset by splitting array
// into K subsets
vector<vector<int> > v(K);
// Size of each subset must
// be less than the number of elements
if (K == 0 || K > N) {
cout << "Not Possible" << endl;
}
else {
cout << "The Subset Combinations are: " << endl;
PartitionSub(arr, 0, N, K, 0, v);
}
}
// Driver Code
int main()
{
// Given array
int arr[] = { 1, 2, 3, 4 };
// Given K
int K = 2;
// Size of the array
int N = sizeof(arr) / sizeof(arr[0]);
// Prints all possible
// splits into subsets
partKSubsets(arr, N, K);
}
Java 语言(一种计算机语言，尤用于创建网站)
// Java program for above approach
import java.util.*;
import java.lang.*;
class Gfg
{
// arr: Store input array
// i: Stores current index of arr
// N: Stores length of arr
// K: Stores count of subsets
// nos: Stores count of feasible subsets formed
// v: Store K subsets of the given array
// Utility function to find all possible
// ways to split array into K subsets
static void PartitionSub(int arr[], int i,
int N, int K, int nos,
ArrayList<ArrayList<Integer>> v)
{
// If count of elements in K subsets
// are greater than or equal to N
if (i >= N)
{
// If count of subsets
// formed is equal to K
if (nos == K)
{
// Print K subsets by splitting
// array into K subsets
for (int x = 0; x < v.size(); x++)
{
System.out.print("{ ");
// Print current subset
for (int y = 0; y < v.get(x).size(); y++)
{
System.out.print(v.get(x).get(y));
// If current element is the last
// element of the subset
if (y == v.get(x).size() - 1)
{
System.out.print(" ");
}
// Otherwise
else
{
System.out.print(", ");
}
}
if (x == v.size() - 1)
{
System.out.print("}");
}
else
{
System.out.print("}, ");
}
}
System.out.println();;
}
return;
}
for (int j = 0; j < K; j++)
{
// If any subset is occupied,
// then push the element
// in that first
if (v.get(j).size() > 0)
{
v.get(j).add(arr[i]);
// Recursively do the same
// for remaining elements
PartitionSub(arr, i + 1, N, K, nos, v);
// Backtrack
v.get(j).remove(v.get(j).size()-1);
}
// Otherwise, push it in an empty
// subset and increase the
// subset count by 1
else
{
v.get(j).add(arr[i]);
PartitionSub(arr, i + 1, N, K, nos + 1, v);
v.get(j).remove(v.get(j).size()-1);
// Break to avoid the case of going in
// other empty subsets, if available,
// and forming the same combination
break;
}
}
}
// Function to to find all possible ways to
// split array into K subsets
static void partKSubsets(int arr[], int N, int K)
{
// Stores K subset by splitting array
// into K subsets
ArrayList<ArrayList<Integer>> v = new ArrayList<>();
for(int i = 0; i < K; i++)
v.add(new ArrayList<>());
// Size of each subset must
// be less than the number of elements
if (K == 0 || K > N)
{
System.out.println("Not Possible");
}
else
{
System.out.println("The Subset Combinations are: ");
PartitionSub(arr, 0, N, K, 0, v);
}
}
// Driver function
public static void main (String[] args)
{
// Given array
int arr[] = { 1, 2, 3, 4 };
// Given K
int K = 2;
// Size of the array
int N = arr.length;
// Prints all possible
// splits into subsets
partKSubsets(arr, N, K);
}
}
// This code is contributed by offbeat
Python 3
# Python 3 program for the above approach
# arr: Store input array
# i: Stores current index of arr
# N: Stores length of arr
# K: Stores count of subsets
# nos: Stores count of feasible subsets formed
# v: Store K subsets of the given array
# Utility function to find all possible
# ways to split array into K subsets
def PartitionSub(arr, i, N, K, nos, v):
# If count of elements in K subsets
# are greater than or equal to N
if (i >= N):
# If count of subsets
# formed is equal to K
if (nos == K):
# Print K subsets by splitting
# array into K subsets
for x in range(len(v)):
print("{ ", end = "")
# Print current subset
for y in range(len(v[x])):
print(v[x][y], end = "")
# If current element is the last
# element of the subset
if (y == len(v[x]) - 1):
print(" ", end = "")
# Otherwise
else:
print(", ", end = "")
if (x == len(v) - 1):
print("}", end = "")
else:
print("}, ", end = "")
print("\n", end = "")
return
for j in range(K):
# If any subset is occupied,
# then push the element
# in that first
if (len(v[j]) > 0):
v[j].append(arr[i])
# Recursively do the same
# for remaining elements
PartitionSub(arr, i + 1, N, K, nos, v)
# Backtrack
v[j].remove(v[j][len(v[j]) - 1])
# Otherwise, push it in an empty
# subset and increase the
# subset count by 1
else:
v[j].append(arr[i])
PartitionSub(arr, i + 1, N, K, nos + 1, v)
v[j].remove(v[j][len(v[j]) - 1])
# Break to avoid the case of going in
# other empty subsets, if available,
# and forming the same combination
break
# Function to to find all possible ways to
# split array into K subsets
def partKSubsets(arr, N, K):
# Stores K subset by splitting array
# into K subsets
v = [[] for i in range(K)]
# Size of each subset must
# be less than the number of elements
if (K == 0 or K > N):
print("Not Possible", end = "")
else:
print("The Subset Combinations are: ")
PartitionSub(arr, 0, N, K, 0, v)
# Driver Code
if __name__ == '__main__':
# Given array
arr = [1, 2, 3, 4]
# Given K
K = 2
# Size of the array
N = len(arr)
# Prints all possible
# splits into subsets
partKSubsets(arr, N, K)
# This code is contributed by bgangwar59.
C
// C# program for the above approach
using System;
using System.Collections.Generic;
class GFG
{
// arr: Store input array
// i: Stores current index of arr
// N: Stores length of arr
// K: Stores count of subsets
// nos: Stores count of feasible subsets formed
// v: Store K subsets of the given array
// Utility function to find all possible
// ways to split array into K subsets
static void PartitionSub(int []arr, int i,
int N, int K, int nos,
List<List<int>>v)
{
// If count of elements in K subsets
// are greater than or equal to N
if (i >= N) {
// If count of subsets
// formed is equal to K
if (nos == K) {
// Print K subsets by splitting
// array into K subsets
for (int x = 0; x < v.Count; x++) {
Console.Write("{ ");
// Print current subset
for (int y = 0; y < v[x].Count; y++) {
Console.Write(v[x][y]);
// If current element is the last
// element of the subset
if (y == v[x].Count - 1) {
Console.Write(" ");
}
// Otherwise
else {
Console.Write(", ");
}
}
if (x == v.Count - 1) {
Console.Write("}");
}
else {
Console.Write("}, ");
}
}
Console.Write("\n");
}
return;
}
for (int j = 0; j < K; j++) {
// If any subset is occupied,
// then push the element
// in that first
if (v[j].Count > 0) {
v[j].Add(arr[i]);
// Recursively do the same
// for remaining elements
PartitionSub(arr, i + 1, N, K, nos, v);
// Backtrack
v[j].RemoveAt(v[j].Count - 1);
}
// Otherwise, push it in an empty
// subset and increase the
// subset count by 1
else {
v[j].Add(arr[i]);
PartitionSub(arr, i + 1, N, K, nos + 1, v);
v[j].RemoveAt(v[j].Count - 1);
// Break to avoid the case of going in
// other empty subsets, if available,
// and forming the same combination
break;
}
}
}
// Function to to find all possible ways to
// split array into K subsets
static void partKSubsets(int []arr, int N, int K)
{
// Stores K subset by splitting array
// into K subsets
List<List<int> > v = new List<List<int>>();
for(int i=0;i<K;i++)
v.Add(new List<int>());
// Size of each subset must
// be less than the number of elements
if (K == 0 || K > N) {
Console.WriteLine("Not Possible");
}
else {
Console.WriteLine("The Subset Combinations are: ");
PartitionSub(arr, 0, N, K, 0, v);
}
}
// Driver Code
public static void Main()
{
// Given array
int []arr = {1, 2, 3, 4};
// Given K
int K = 2;
// Size of the array
int N = arr.Length;
// Prints all possible
// splits into subsets
partKSubsets(arr, N, K);
}
}
// This code is contributed by SURENDRA_GANGWAR.
java 描述语言
<script>
// Javascript program for above approach
// arr: Store input array
// i: Stores current index of arr
// N: Stores length of arr
// K: Stores count of subsets
// nos: Stores count of feasible subsets formed
// v: Store K subsets of the given array
// Utility function to find all possible
// ways to split array into K subsets
function PartitionSub(arr, i, N, K, nos, v)
{
// If count of elements in K subsets
// are greater than or equal to N
if (i >= N)
{
// If count of subsets
// formed is equal to K
if (nos == K)
{
// Print K subsets by splitting
// array into K subsets
for (let x = 0; x < v.length; x++)
{
document.write("{ ");
// Print current subset
for (let y = 0; y < v[x].length; y++)
{
document.write(v[x][y]);
// If current element is the last
// element of the subset
if (y == v[x].length - 1)
{
document.write(" ");
}
// Otherwise
else
{
document.write(", ");
}
}
if (x == v.length - 1)
{
document.write("}");
}
else
{
document.write("}, ");
}
}
document.write("</br>");
}
return;
}
for (let j = 0; j < K; j++)
{
// If any subset is occupied,
// then push the element
// in that first
if (v[j].length > 0)
{
v[j].push(arr[i]);
// Recursively do the same
// for remaining elements
PartitionSub(arr, i + 1, N, K, nos, v);
// Backtrack
v[j].pop();
}
// Otherwise, push it in an empty
// subset and increase the
// subset count by 1
else
{
v[j].push(arr[i]);
PartitionSub(arr, i + 1, N, K, nos + 1, v);
v[j].pop();
// Break to avoid the case of going in
// other empty subsets, if available,
// and forming the same combination
break;
}
}
}
// Function to to find all possible ways to
// split array into K subsets
function partKSubsets(arr, N, K)
{
// Stores K subset by splitting array
// into K subsets
let v = [];
for(let i = 0; i < K; i++)
v.push([]);
// Size of each subset must
// be less than the number of elements
if (K == 0 || K > N)
{
document.write("Not Possible" + "</br>");
}
else
{
document.write("The Subset Combinations are: " + "</br>");
PartitionSub(arr, 0, N, K, 0, v);
}
}
// Given array
let arr = [ 1, 2, 3, 4 ];
// Given K
let K = 2;
// Size of the array
let N = arr.length;
// Prints all possible
// splits into subsets
partKSubsets(arr, N, K);
// This code is contributed by decode2207.
</script>
Output
The Subset Combinations are:
{ 1, 2, 3 }, { 4 }
{ 1, 2, 4 }, { 3 }
{ 1, 2 }, { 3, 4 }
{ 1, 3, 4 }, { 2 }
{ 1, 3 }, { 2, 4 }
{ 1, 4 }, { 2, 3 }
{ 1 }, { 2, 3, 4 }
版权属于：月萌API www.moonapi.com，转载请注明出处
