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📜  来自n个数组的组合从每个数组中选择一个元素

📅  最后修改于: 2021-06-25 13:13:06             🧑  作者: Mango

给定一个数组列表,找到所有组合,其中每个组合包含每个给定数组中的一个元素。
例子:

Input : [ [1, 2], [3, 4] ]
Output : 1 3  
         1 4 
         2 3 
         2 4       

Input : [ [1], [2, 3, 4], [5] ]
Output : 1 2 5  
         1 3 5
         1 4 5           

我们保持大小等于数组总数的数组。这个称为索引的数组可帮助我们跟踪n个数组中每个数组中当前元素的索引。最初,它用全0初始化,表示每个数组中的当前索引是第一个元素的索引。我们会继续打印组合,直到找不到新的组合。从最右边的数组开始,我们检查该数组中是否还有更多元素。如果是,我们将以索引的形式增加该数组的条目,即移至该数组中的下一个元素。我们还使该数组右边所有数组中的当前索引为0。我们一直向左移动以检查所有阵列,直到找到一个这样的阵列。如果找不到更多的数组,我们就到此为止。

C++
// C++ program to find combinations from n
// arrays such that one element from each
// array is present
#include 
 
using namespace std;
 
// function to print combinations that contain
// one element from each of the given arrays
void print(vector >& arr)
{
    // number of arrays
    int n = arr.size();
 
    // to keep track of next element in each of
    // the n arrays
    int* indices = new int[n];
 
    // initialize with first element's index
    for (int i = 0; i < n; i++)
        indices[i] = 0;
 
    while (1) {
 
        // print current combination
        for (int i = 0; i < n; i++)
            cout << arr[i][indices[i]] << " ";
        cout << endl;
 
        // find the rightmost array that has more
        // elements left after the current element
        // in that array
        int next = n - 1;
        while (next >= 0 &&
              (indices[next] + 1 >= arr[next].size()))
            next--;
 
        // no such array is found so no more
        // combinations left
        if (next < 0)
            return;
 
        // if found move to next element in that
        // array
        indices[next]++;
 
        // for all arrays to the right of this
        // array current index again points to
        // first element
        for (int i = next + 1; i < n; i++)
            indices[i] = 0;
    }
}
 
// driver function to test above function
int main()
{
    // initializing a vector with 3 empty vectors
    vector > arr(3, vector(0, 0));
 
    // now entering data
    // [[1, 2, 3], [4], [5, 6]]
    arr[0].push_back(1);
    arr[0].push_back(2);
    arr[0].push_back(3);
    arr[1].push_back(4);
    arr[2].push_back(5);
    arr[2].push_back(6);
 
    print(arr);
}


Java
// Java program to find combinations from n
// arrays such that one element from each
// array is present
import java.util.*;
 
class GFG{
 
// Function to print combinations that contain
// one element from each of the given arrays
static void print(Vector []arr)
{
     
    // Number of arrays
    int n = arr.length;
 
    // To keep track of next element in
    // each of the n arrays
    int []indices = new int[n];
 
    // Initialize with first element's index
    for(int i = 0; i < n; i++)
        indices[i] = 0;
 
    while (true)
    {
 
        // Print current combination
        for(int i = 0; i < n; i++)
            System.out.print(
                arr[i].get(indices[i]) + " ");
                 
        System.out.println();
 
        // Find the rightmost array that has more
        // elements left after the current element
        // in that array
        int next = n - 1;
        while (next >= 0 &&
              (indices[next] + 1 >=
                   arr[next].size()))
            next--;
 
        // No such array is found so no more
        // combinations left
        if (next < 0)
            return;
 
        // If found move to next element in that
        // array
        indices[next]++;
 
        // For all arrays to the right of this
        // array current index again points to
        // first element
        for(int i = next + 1; i < n; i++)
            indices[i] = 0;
    }
}
 
// Driver code
public static void main(String[] args)
{
     
    // Initializing a vector with 3 empty vectors
    @SuppressWarnings("unchecked")
    Vector []arr = new Vector[3];
    for(int i = 0; i < arr.length; i++)
        arr[i] = new Vector();
         
    // Now entering data
    // [[1, 2, 3], [4], [5, 6]]
    arr[0].add(1);
    arr[0].add(2);
    arr[0].add(3);
    arr[1].add(4);
    arr[2].add(5);
    arr[2].add(6);
 
    print(arr);
}
}
 
// This code is contributed by amal kumar choubey


Python3
# Python3 program to find combinations from n
# arrays such that one element from each
# array is present
 
# function to prcombinations that contain
# one element from each of the given arrays
def print1(arr):
     
    # number of arrays
    n = len(arr)
 
    # to keep track of next element
    # in each of the n arrays
    indices = [0 for i in range(n)]
 
    while (1):
 
        # prcurrent combination
        for i in range(n):
            print(arr[i][indices[i]], end = " ")
        print()
 
        # find the rightmost array that has more
        # elements left after the current element
        # in that array
        next = n - 1
        while (next >= 0 and
              (indices[next] + 1 >= len(arr[next]))):
            next-=1
 
        # no such array is found so no more
        # combinations left
        if (next < 0):
            return
 
        # if found move to next element in that
        # array
        indices[next] += 1
 
        # for all arrays to the right of this
        # array current index again points to
        # first element
        for i in range(next + 1, n):
            indices[i] = 0
 
# Driver Code
 
# initializing a vector with 3 empty vectors
arr = [[] for i in range(3)]
 
# now entering data
# [[1, 2, 3], [4], [5, 6]]
arr[0].append(1)
arr[0].append(2)
arr[0].append(3)
arr[1].append(4)
arr[2].append(5)
arr[2].append(6)
 
print1(arr)
 
# This code is contributed by mohit kumar


C#
// C# program to find
// combinations from n
// arrays such that one
// element from each
// array is present
using System;
using System.Collections.Generic;
class GFG{
 
// Function to print combinations
// that contain one element from
// each of the given arrays
static void print(List []arr)
{
  // Number of arrays
  int n = arr.Length;
 
  // To keep track of next
  // element in each of
  // the n arrays
  int []indices = new int[n];
 
  // Initialize with first
  // element's index
  for(int i = 0; i < n; i++)
    indices[i] = 0;
 
  while (true)
  {
    // Print current combination
    for(int i = 0; i < n; i++)
      Console.Write(arr[i][indices[i]] + " ");
     
    Console.WriteLine();
 
    // Find the rightmost array
    // that has more elements
    // left after the current
    // element in that array
    int next = n - 1;
    while (next >= 0 &&
          (indices[next] + 1 >=
           arr[next].Count))
      next--;
 
    // No such array is found
    // so no more combinations left
    if (next < 0)
      return;
 
    // If found move to next
    // element in that array
    indices[next]++;
 
    // For all arrays to the right
    // of this array current index
    // again points to first element
    for(int i = next + 1; i < n; i++)
      indices[i] = 0;
  }
}
 
// Driver code
public static void Main(String[] args)
{
  // Initializing a vector
  // with 3 empty vectors
  List []arr = new List[3];
  for(int i = 0; i < arr.Length; i++)
    arr[i] = new List();
 
  // Now entering data
  // [[1, 2, 3], [4], [5, 6]]
  arr[0].Add(1);
  arr[0].Add(2);
  arr[0].Add(3);
  arr[1].Add(4);
  arr[2].Add(5);
  arr[2].Add(6);
 
  print(arr);
}
}
 
// This code is contributed by shikhasingrajput


输出:

1 4 5
1 4 6
2 4 5
2 4 6
3 4 5
3 4 6