给定一个大小为N的数组arr[] ，任务是生成并打印数组中R元素的所有可能组合。
例子：
Input: arr[] = {0, 1, 2, 3}, R = 3
Output:
0 1 2
0 1 3
0 2 3
1 2 3
Input: arr[] = {1, 3, 4, 5, 6, 7}, R = 5
Output:
1 3 4 5 6
1 3 4 5 7
1 3 4 6 7
1 3 5 6 7
1 4 5 6 7
3 4 5 6 7
方法：这里讨论递归方法。在这篇文章中，将讨论输出给定数组的所有组合的迭代方法。
迭代方法充当状态机。当机器被调用时，它输出一个组合并移动到下一个。
对于大小为n的数组中的r 个元素的组合，可以在组合中包括或排除给定元素。
让我们有一个大小为n的布尔数组来标记是否包含数据数组中的相应元素。如果数据阵列中的第i个元素被包括，则布尔阵列中的第i个元素是真还是假，否则。
然后，布尔数组中的r 个布尔值将被标记为 true。我们可以初始化布尔数组，使其从索引0到索引r – 1有r 个真值。在迭代过程中，我们从左到右扫描布尔数组，找到第一个为真且前一个为假的元素，以及第一个为真且下一个为假的元素。
然后，我们在布尔数组中有第一个连续的真值。假设在这个区域有m 个真值，从索引Start 开始到索引End 结束。下一次迭代将是
- 将布尔数组的索引End + 1设置为 true。
- 将 index Start到 index End – 1的布尔数组设置为 false。
- 将索引0到索引k – 2 设置为 true。
例如，
如果当前布尔数组是{0, 0, 1, 1, 1, 1, 0, 0, 0, 1, 0, 0} ，则k = 4 ， Start = 2 ， End = 5 。下一个布尔数组将是{1, 1, 1, 0, 0, 0, 1, 0, 0, 1, 0, 0} 。如果Start == End 区域中只有一个 true，我们只需将索引End设置为 false 并将索引End + 1 设置为 true。
我们还需要记录当前的Start和End ，并在每次迭代期间更新Start和End 。当最后r 个布尔值设置为 true 时，我们无法移动到下一个组合，我们停止。
下图说明了布尔数组如何从一次迭代更改为另一次迭代。
要输出组合，我们只需扫描布尔数组。如果它的第i个索引为真，我们打印出数据数组的第i个元素。
下面是上述方法的实现：
C++
// C++ implementation of the approach
#include
using namespace std;
class Combination {
private:
// Data array for combination
int* Indices;
// Length of the data array
int N;
// Number of elements in the combination
int R;
// The boolean array
bool* Flags;
// Starting index of the 1st tract of trues
int Start;
// Ending index of the 1st tract of trues
int End;
public:
// Constructor
Combination(int* arr, int n, int r)
{
this->Indices = arr;
this->N = n;
this->R = r;
this->Flags = nullptr;
}
~Combination()
{
if (this->Flags != nullptr) {
delete[] this->Flags;
}
}
// Set the 1st r Booleans to true,
// initialize Start and End
void GetFirst()
{
this->Flags = new bool[N];
// Generate the very first combination
for (int i = 0; i < this->N; ++i) {
if (i < this->R) {
Flags[i] = true;
}
else {
Flags[i] = false;
}
}
// Update the starting ending indices
// of trues in the boolean array
this->Start = 0;
this->End = this->R - 1;
this->Output();
}
// Function that returns true if another
// combination can still be generated
bool HasNext()
{
return End < (this->N - 1);
}
// Function to generate the next combination
void Next()
{
// Only one true in the tract
if (this->Start == this->End) {
this->Flags[this->End] = false;
this->Flags[this->End + 1] = true;
this->Start += 1;
this->End += 1;
while (this->End + 1 < this->N
&& this->Flags[this->End + 1]) {
++this->End;
}
}
else {
// Move the End and reset the End
if (this->Start == 0) {
Flags[this->End] = false;
Flags[this->End + 1] = true;
this->End -= 1;
}
else {
Flags[this->End + 1] = true;
// Set all the values to false starting from
// index Start and ending at index End
// in the boolean array
for (int i = this->Start; i <= this->End; ++i) {
Flags[i] = false;
}
// Set the beginning elements to true
for (int i = 0; i < this->End - this->Start; ++i) {
Flags[i] = true;
}
// Reset the End
this->End = this->End - this->Start - 1;
this->Start = 0;
}
}
this->Output();
}
private:
// Function to print the combination generated previouslt
void Output()
{
for (int i = 0, count = 0; i < this->N
&& count < this->R;
++i) {
// If current index is set to true in the boolean array
// then element at current index in the original array
// is part of the combination generated previously
if (Flags[i]) {
cout << Indices[i] << " ";
++count;
}
}
cout << endl;
}
};
// Driver code
int main()
{
int arr[] = { 0, 1, 2, 3 };
int n = sizeof(arr) / sizeof(int);
int r = 3;
Combination com(arr, n, r);
com.GetFirst();
while (com.HasNext()) {
com.Next();
}
return 0;
} Java
// Java implementation of the approach
class Combination
{
// Data array for combination
private int[] Indices;
// Number of elements in the combination
private int R;
// The boolean array
private boolean[] Flags;
// Starting index of the 1st tract of trues
private int Start;
// Ending index of the 1st tract of trues
private int End;
// Constructor
public Combination(int[] arr, int r)
{
this.Indices = arr;
this.R = r;
}
// Set the 1st r Booleans to true,
// initialize Start and End
public void GetFirst()
{
Flags = new boolean[this.Indices.length];
// Generate the very first combination
for (int i = 0; i < this.R; ++i)
{
Flags[i] = true;
}
// Update the starting ending indices
// of trues in the boolean array
this.Start = 0;
this.End = this.R - 1;
this.Output();
}
// Function that returns true if another
// combination can still be generated
public boolean HasNext()
{
return End < (this.Indices.length - 1);
}
// Function to generate the next combination
public void Next()
{
// Only one true in the tract
if (this.Start == this.End)
{
this.Flags[this.End] = false;
this.Flags[this.End + 1] = true;
this.Start += 1;
this.End += 1;
while (this.End + 1 < this.Indices.length
&& this.Flags[this.End + 1])
{
++this.End;
}
}
else
{
// Move the End and reset the End
if (this.Start == 0)
{
Flags[this.End] = false;
Flags[this.End + 1] = true;
this.End -= 1;
}
else
{
Flags[this.End + 1] = true;
// Set all the values to false starting from
// index Start and ending at index End
// in the boolean array
for (int i = this.Start; i <= this.End; ++i)
{
Flags[i] = false;
}
// Set the beginning elements to true
for (int i = 0; i < this.End - this.Start; ++i)
{
Flags[i] = true;
}
// Reset the End
this.End = this.End - this.Start - 1;
this.Start = 0;
}
}
this.Output();
}
// Function to print the combination generated previouslt
private void Output()
{
for (int i = 0, count = 0; i < Indices.length
&& count < this.R; ++i)
{
// If current index is set to true in the boolean array
// then element at current index in the original array
// is part of the combination generated previously
if (Flags[i])
{
System.out.print(Indices[i]);
System.out.print(" ");
++count;
}
}
System.out.println();
}
}
// Driver code
class GFG
{
public static void main(String[] args)
{
int[] arr = { 0, 1, 2, 3 };
int r = 3;
Combination com = new Combination(arr, r);
com.GetFirst();
while (com.HasNext())
{
com.Next();
}
}
}
// This code is contributed by Rajput-JiC#
// C# implementation of the approach
using System;
namespace IterativeCombination {
class Combination {
// Data array for combination
private int[] Indices;
// Number of elements in the combination
private int R;
// The boolean array
private bool[] Flags;
// Starting index of the 1st tract of trues
private int Start;
// Ending index of the 1st tract of trues
private int End;
// Constructor
public Combination(int[] arr, int r)
{
this.Indices = arr;
this.R = r;
}
// Set the 1st r Booleans to true,
// initialize Start and End
public void GetFirst()
{
Flags = new bool[this.Indices.Length];
// Generate the very first combination
for (int i = 0; i < this.R; ++i) {
Flags[i] = true;
}
// Update the starting ending indices
// of trues in the boolean array
this.Start = 0;
this.End = this.R - 1;
this.Output();
}
// Function that returns true if another
// combination can still be generated
public bool HasNext()
{
return End < (this.Indices.Length - 1);
}
// Function to generate the next combination
public void Next()
{
// Only one true in the tract
if (this.Start == this.End) {
this.Flags[this.End] = false;
this.Flags[this.End + 1] = true;
this.Start += 1;
this.End += 1;
while (this.End + 1 < this.Indices.Length
&& this.Flags[this.End + 1]) {
++this.End;
}
}
else {
// Move the End and reset the End
if (this.Start == 0) {
Flags[this.End] = false;
Flags[this.End + 1] = true;
this.End -= 1;
}
else {
Flags[this.End + 1] = true;
// Set all the values to false starting from
// index Start and ending at index End
// in the boolean array
for (int i = this.Start; i <= this.End; ++i) {
Flags[i] = false;
}
// Set the beginning elements to true
for (int i = 0; i < this.End - this.Start; ++i) {
Flags[i] = true;
}
// Reset the End
this.End = this.End - this.Start - 1;
this.Start = 0;
}
}
this.Output();
}
// Function to print the combination generated previouslt
private void Output()
{
for (int i = 0, count = 0; i < Indices.Length
&& count < this.R;
++i) {
// If current index is set to true in the boolean array
// then element at current index in the original array
// is part of the combination generated previously
if (Flags[i]) {
Console.Write(Indices[i]);
Console.Write(" ");
++count;
}
}
Console.WriteLine();
}
}
// Driver code
class AppDriver {
static void Main()
{
int[] arr = { 0, 1, 2, 3 };
int r = 3;
Combination com = new Combination(arr, r);
com.GetFirst();
while (com.HasNext()) {
com.Next();
}
}
}
}输出：
0 1 2
0 1 3
0 2 3
1 2 3
如果您希望与专家一起参加现场课程，请参阅DSA 现场工作专业课程和学生竞争性编程现场课程。