Question

我需要一些关于如何使用数组来打印所有可能序列的想法。例如，

array 1: AA BB
array 2: CC
array 3: DD EE FF
array 4: GG

现在我需要列出任何给定数组的所有可能组合，每个数组只使用1个序列，如下所示：

AA CC DD GG
AA CC EE GG
AA CC FF GG
BB CC DD GG
BB CC EE GG
BB CC FF GG

有没有人知道或者可以让我开始讨论如何做到这一点？

Answer 1

如果这些是4个不同的数组，我想不出更好的选择，然后编写4个嵌套循环，每个迭代遍历其中一个数组。如果你有一个包含所有数组的二维数组，我建议你使用递归。

Answer 2

我认为你的意思是每个数组的元素数量是未知的，我使用了sizeof（）。和其他人一样，你只需要为循环嵌套5个。

int main()
{
    //naming arrays a,b,c,d,e, change accordingly
    //this function prints the combinations of 1 char
    int aElements = sizeof(a) / sizeof(a[0]);
    int bElements = sizeof(b) / sizeof(b[0]);
    int cElements = sizeof(c) / sizeof(c[0]);
    int dElements = sizeof(d) / sizeof(d[0]);
    int eElements = sizeof(e) / sizeof(e[0]);
    for (int i = 0; i < aElements; i++){
        for (int j = 0; j < bElements; j++){
            for (int k = 0; k < cElements; k++){
                for (int l = 0; l < dElements; l++){
                    for (int m = 0; m < eElements; m++){
                        cout << a[i] << b[j] << c[k] << d[l] << e[m] << endl;
                    }
                }
            }
        }
    }
}

为了找出组合的数量，您可以将计数器放在内部循环中，或者只是将元素数除以组合数（在本例中为1）并将它们全部相乘。 E.G，在您的示例中，它将是（4/1）*（2/1）*（2/1）*（6/1）*（2/1）= 192种组合。如果你每2个字符组合，在你的第二个例子中它将是（4/2）*（2/2）*（2/2）*（6/2）*（2/2）= 6种组合。以下功能打印出2的组合。

int main()
    {
    //naming arrays a,b,c,d,e, change accordingly
    //this function prints the combinations of 2 chars
    int aElements = sizeof(a) / sizeof(a[0]);
    int bElements = sizeof(b) / sizeof(b[0]);
    int cElements = sizeof(c) / sizeof(c[0]);
    int dElements = sizeof(d) / sizeof(d[0]);
    int eElements = sizeof(e) / sizeof(e[0]);
    for (int i = 0; i < aElements - 1; i+=2){
        for (int j = 0; j < bElements - 1; j+=2){
            for (int k = 0; k < cElements - 1; k+=2){
                for (int l = 0; l < dElements - 1; l+=2){
                    for (int m = 0; m < eElements - 1; m+=2){
                        cout << a[i] << a[i+1] << b[j] << b[j+1] << c[k] << c[k+1]  << d[l] << d[l+1] << e[m] << e[m+1] << endl;
                        }
                    }
                }
            }
        }
}

我为第二个做的是增量计数器2而不是1，从元素数减去1不要超出界限，并打印2个连续元素而不是1.这将适用于任意数量的字符组合。

Answer 3

C ++ 11风格！

#include <iostream>
#include <vector>
#include <utility>
#include <iterator>

// metaprogramming boilerplate:

template<typename... L>
struct first_type {};
template<typename T, typename... L>
struct first_type<T, L...> {
  typedef T type;
};
template<typename... L>
using FirstType = typename first_type<L...>::type;

namespace aux {
    using std::begin;
    template<typename C>
    auto adl_begin( C&&c )->decltype( begin(std::forward<C>(c)) );
    template<typename C>
    auto adl_cbegin( C const&c )->decltype( begin(c) );
}
template<typename Container>
struct iterator_type {
  typedef decltype( aux::adl_begin(std::declval<Container>()) ) iterator;
  typedef decltype( aux::adl_cbegin(std::declval<Container>()) ) const_iterator;
};
template<typename Container>
using IteratorType = typename iterator_type<Container>::iterator;

template<typename Container>
struct value_type {
  typedef typename std::iterator_traits< IteratorType<Container> >::value_type type;
};
template<typename Container>
using ValueType = typename value_type<Container>::type;

// Actual problem specific code:
template<typename Func, typename T>
void ForEachPossibility_Helper( Func&& f, std::vector<T>& result) {
  f(result);
}

template<typename Func, typename T, typename Container, typename... Containers>
void ForEachPossibility_Helper( Func&& f, std::vector<T>& result, Container&& arr0, Containers&&... arrays) {
  for( auto const& str:arr0 ) {
    result.push_back(str);
    ForEachPossibility_Helper( std::forward<Func>(f), result, std::forward<Containers>(arrays)... );
    result.pop_back();
  }
}

template<typename Func, typename... Containers>
void ForEachPossibility( Func&& f, Containers&&... arrays) {
    typedef ValueType<FirstType<Containers...>> T;
    std::vector<T> result;
    ForEachPossibility_Helper( std::forward<Func>(f), result, std::forward<Containers>(arrays)... );
}

const char* arr1[] = {"AA", "BB"};
const char* arr2[] = {"CC"};
const char* arr3[] = {"DD", "EE", "FF"};
const char* arr4[] = {"GG"};

int main() {
  ForEachPossibility( []( std::vector<const char*> const& result ){
    for( auto&& str:result ) {
      std::cout << str;
    }
    std::cout << "\n";
  }, arr1, arr2, arr3, arr4 );
}

请注意，只有2个for循环，其中一个用于打印。

Answer 4

据我所知，你不需要关心你从中得到序列的数组的顺序。在这种情况下，递归确实很有用。看起来像这样：

void printSequences(ListOfYourArrays list, int index) {
    if (list.size() > index) {
        array a = list.getElementAt(index);
        //Make a cycle that reads items from your array one by one
        while (...)
            System.out.print(item);
        //And now you need to print all combinations for the rest of arrays in you list
        printSequences(list, index + 1);
    } else
        System.out.println();
}

您需要做的就是将您的数组添加到列表中并调用函数

printSequences(list, 0);

Answer 5

编辑更新

我们需要而不是将每个indice更新为1，而是通过迭代组合来更新......

请参阅：How can I iterate throught every possible combination of n playing cards

现在它看起来像这样

#include <iostream>
#include <vector>
#include <string>
using namespace std;

bool UpdateCombination (std::vector<int> &comboindices, int count, int n)
{
    for (int i = 1; i <= n; ++i)
    {
        if (comboindices[n - i] < count - i)
        {
            ++comboindices[n - i];
            for (int j = n - i + 1; j < n; ++j)
            {
                comboindices[j] = comboindices[j-1] + 1;
            }
            return false;
        }
    }
    return true;
}

void ResetCombination (std::vector<int> &comboindices, int n)
{
    comboindices.resize(n);
    for (int i = 0; i < n; ++i)
    {
        comboindices[i] = i;
    }
}

void PrintArrays (const std::vector<std::vector<std::string>> items, int count)
{
    std::vector<std::vector<int>> indices;
    int n = items.size();
    indices.resize(items.size());

    for(auto i = indices.begin (); i != indices.end (); ++i)
    {
        ResetCombination((*i),count);
    }

    while (true) //Iterate until we've used all of the last array of items
    {
            for (int i = 0; i < n; ++i)
            {
                cout << "{";
                for (auto j = indices[i].begin (); j != indices[i].end (); ++j)
                {
                    int ji = (*j);
                    cout << (items[i])[ji] << " ";
                }
                cout << "} ";

            }
            cout << endl;

            //Update to the next indice
            for (int i = n - 1; i >= 0; --i)
            {
                    bool done = UpdateCombination (indices[i],items[i].size(),count);
                    if (!done)
                    {
                            break;
                    }
                    else if (done && i == 0)
                    {
                        return; //Escape.
                    }
                    else
                    {
                        ResetCombination(indices[i],count);
                    }
            }
    }

}
 //{A,B,C,D},{A,B},{A,B},{A,B,C,D,E,F},{A,B}


int main() {

    vector<vector<string>> lists;
    lists.resize(5);
    lists[0].push_back("A");
    lists[0].push_back("B");
    lists[0].push_back("C");
    lists[0].push_back("D");

    lists[1].push_back("A");
    lists[1].push_back("B");

    lists[2].push_back("A");
    lists[2].push_back("B");

    lists[3].push_back("A");
    lists[3].push_back("B");
    lists[3].push_back("C");
    lists[3].push_back("D");
    lists[3].push_back("E");
    lists[3].push_back("F");

    lists[4].push_back("A");
    lists[4].push_back("B");



    PrintArrays(lists,2);

    int pause;
    cin >> pause;
    return 0;
}

给我们......

{A B } {A B } {A B } {A B } {A B } 
{A B } {A B } {A B } {A C } {A B } 
{A B } {A B } {A B } {A D } {A B } 
{A B } {A B } {A B } {A E } {A B } 
{A B } {A B } {A B } {A F } {A B } 
{A B } {A B } {A B } {B C } {A B } 
{A B } {A B } {A B } {B D } {A B } 
{A B } {A B } {A B } {B E } {A B } 
{A B } {A B } {A B } {B F } {A B } 
{A B } {A B } {A B } {C D } {A B } 
{A B } {A B } {A B } {C E } {A B } 
{A B } {A B } {A B } {C F } {A B } 
{A B } {A B } {A B } {D E } {A B } 
{A B } {A B } {A B } {D F } {A B } 
{A B } {A B } {A B } {E F } {A B } 
{A C } {A B } {A B } {A B } {A B } 
{A C } {A B } {A B } {A C } {A B } 
{A C } {A B } {A B } {A D } {A B } 
{A C } {A B } {A B } {A E } {A B } 
{A C } {A B } {A B } {A F } {A B } 
{A C } {A B } {A B } {B C } {A B } 
{A C } {A B } {A B } {B D } {A B } 
{A C } {A B } {A B } {B E } {A B } 
{A C } {A B } {A B } {B F } {A B } 
{A C } {A B } {A B } {C D } {A B } 
{A C } {A B } {A B } {C E } {A B } 
{A C } {A B } {A B } {C F } {A B } 
{A C } {A B } {A B } {D E } {A B } 
{A C } {A B } {A B } {D F } {A B } 
{A C } {A B } {A B } {E F } {A B } 
{A D } {A B } {A B } {A B } {A B } 
{A D } {A B } {A B } {A C } {A B } 
{A D } {A B } {A B } {A D } {A B } 
{A D } {A B } {A B } {A E } {A B } 
{A D } {A B } {A B } {A F } {A B } 
{A D } {A B } {A B } {B C } {A B } 
{A D } {A B } {A B } {B D } {A B } 
{A D } {A B } {A B } {B E } {A B } 
{A D } {A B } {A B } {B F } {A B } 
{A D } {A B } {A B } {C D } {A B } 
{A D } {A B } {A B } {C E } {A B } 
{A D } {A B } {A B } {C F } {A B } 
{A D } {A B } {A B } {D E } {A B } 
{A D } {A B } {A B } {D F } {A B } 
{A D } {A B } {A B } {E F } {A B } 
{B C } {A B } {A B } {A B } {A B } 
{B C } {A B } {A B } {A C } {A B } 
{B C } {A B } {A B } {A D } {A B } 
{B C } {A B } {A B } {A E } {A B } 
{B C } {A B } {A B } {A F } {A B } 
{B C } {A B } {A B } {B C } {A B } 
{B C } {A B } {A B } {B D } {A B } 
{B C } {A B } {A B } {B E } {A B } 
{B C } {A B } {A B } {B F } {A B } 
{B C } {A B } {A B } {C D } {A B } 
{B C } {A B } {A B } {C E } {A B } 
{B C } {A B } {A B } {C F } {A B } 
{B C } {A B } {A B } {D E } {A B } 
{B C } {A B } {A B } {D F } {A B } 
{B C } {A B } {A B } {E F } {A B } 
{B D } {A B } {A B } {A B } {A B } 
{B D } {A B } {A B } {A C } {A B } 
{B D } {A B } {A B } {A D } {A B } 
{B D } {A B } {A B } {A E } {A B } 
{B D } {A B } {A B } {A F } {A B } 
{B D } {A B } {A B } {B C } {A B } 
{B D } {A B } {A B } {B D } {A B } 
{B D } {A B } {A B } {B E } {A B } 
{B D } {A B } {A B } {B F } {A B } 
{B D } {A B } {A B } {C D } {A B } 
{B D } {A B } {A B } {C E } {A B } 
{B D } {A B } {A B } {C F } {A B } 
{B D } {A B } {A B } {D E } {A B } 
{B D } {A B } {A B } {D F } {A B } 
{B D } {A B } {A B } {E F } {A B } 
{C D } {A B } {A B } {A B } {A B } 
{C D } {A B } {A B } {A C } {A B } 
{C D } {A B } {A B } {A D } {A B } 
{C D } {A B } {A B } {A E } {A B } 
{C D } {A B } {A B } {A F } {A B } 
{C D } {A B } {A B } {B C } {A B } 
{C D } {A B } {A B } {B D } {A B } 
{C D } {A B } {A B } {B E } {A B } 
{C D } {A B } {A B } {B F } {A B } 
{C D } {A B } {A B } {C D } {A B } 
{C D } {A B } {A B } {C E } {A B } 
{C D } {A B } {A B } {C F } {A B } 
{C D } {A B } {A B } {D E } {A B } 
{C D } {A B } {A B } {D F } {A B } 
{C D } {A B } {A B } {E F } {A B }

查看输出。 http://ideone.com/L5AZVv

旧的意思链接： http://ideone.com/58ARAZ

Answer 6

如果你有一个可变数量的数组，那么

Tocs的回答是正确的。如果你总是有4个数组，你可以简单地使用4个嵌套循环。

   for (unsigned int i1 = 0; i1 < a1.size(); ++i1)
        for (unsigned int i2 = 0; i2 < a2.size(); ++i2)
            for (unsigned int i3 = 0; i3 < a3.size(); ++i3)
                for (unsigned int i4 = 0; i4 < a4.size(); ++i4)
                    cout << a1[i1] << " " << a2[i2] << " " << a3[i3] << " " << a4[i4] << std::endl;

有关完整代码，请参见此处http://ideone.com/YcW84Q。

Answer 7

另一种可能性是“计算”当前组合（例如，计算二进制）。从（0,0,0,0）开始计算最大数组索引（1,0,2,0）。在每个步骤中，首先将1添加到第一个索引。如果它大于最大索引（此处为1），则将其设置为零并继续下一个索引

结果：

（0,0,0,0） - ＆gt; AA CC DD GG

（1,0,0,0） - ＆gt; BB CC DD GG

（0,0,1,0） - ＆gt; AA CC EE GG

（1,0,1,0） - ＆gt; BB CC EE GG

（0,0,2,0） - ＆gt; AA CC FF GG

（1,0,2,0） - ＆gt; BB CC FF GG

Answer 8

4个循环导致N pow（4）。

将4个数组分成2个循环。

for each(arr 1){
  for each(arr 2)
  insert into container 1.
}


for each(arr 3){
  for each(arr 4)
  insert into container 2.
}


for each(container 1){
  for each(container 2)
  insert into container3 (*iter 1 + *iter 2)
}

因此复杂性将是最大3NPow（2），这将小于N pow（4）

制作用于列出所有可能序列的算法

8 个答案: