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.