如何编写迭代算法生成一组的所有子集？ [英] how to write iterative algorithm for generate all subsets of a set?

问题描述

我写的递归回溯算法找到一个给定的所有子集。

I wrote recursive backtracking algorithm for finding all subsets of a given set.

void backtracke(int* a, int k, int n)
{
    if (k == n)
    {
        for(int i = 1; i <=k; ++i)
        {
            if (a[i] == true)
            {
                std::cout << i << " ";
            }
        }
        std::cout << std::endl;
        return;
    }
    bool c[2];
    c[0] = false;
    c[1] = true;
    ++k;
    for(int i = 0; i < 2; ++i)
    {       
        a[k] = c[i];
        backtracke(a, k, n);
        a[k] = INT_MAX;
    }
}

现在我们必须写同样的算法，但在反复的形式，该怎么办呢？

now we have to write the same algorithm but in an iterative form, how to do it ?

推荐答案

您可以使用二进制计数器的方法。长度为n重新$ P $的任何唯一的二进制串psents一组n个元素的独特的子集。如果你开始用0和2 ^ n-1个结束，您覆盖所有可能的子集。计数器可以以迭代的方式容易地实现。

You can use the binary counter approach. Any unique binary string of length n represents a unique subset of a set of n elements. If you start with 0 and end with 2^n-1, you cover all possible subsets. The counter can be easily implemented in an iterative manner.

在code在Java中：

The code in Java:

  public static void printAllSubsets(int[] arr) {
    byte[] counter = new byte[arr.length];

    while (true) {
      // Print combination
      for (int i = 0; i < counter.length; i++) {
        if (counter[i] != 0)
          System.out.print(arr[i] + " ");
      }
      System.out.println();

      // Increment counter
      int i = 0;
      while (i < counter.length && counter[i] == 1)
        counter[i++] = 0;
      if (i == counter.length)
        break;
      counter[i] = 1;
    }
  }

请注意，在Java中你可以使用位集合，这使得code真短，但我用一个字节数组来说明这个过程更好。

Note that in Java one can use BitSet, which makes the code really shorter, but I used a byte array to illustrate the process better.

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