DFS 查找子集的所有排列

Question

Given a string with no duplicate characters, return a list with all permutations of the string and all its subsets.

Examples

Set = "abc" , all permutations are: ["", "a", "ab", "abc", "ac", "acb", "b", "ba", "bac", "bc", "bca", "c", "cb", "cba", "ca", "cab"] .

I know how to do it if I broke up the whole thing into two parts:

1. find all subsets
2. for each subset, find all permutations

But I am wondering if there is a way to write the DFS helper function to do it in just one step.

Answer 1

Sure. This isn't the fastest method, but it's simple and enumerates in the lexicographic order such that the input string is the minimum permutation.

private static void printPerms(char[] set, int i) {
  System.out.println(new String(set, 0, i));
  for (int j = i; j < set.length; j++) {
    char c = set[j];
    // move c to position i
    for (int k = j; k > i; k--) {
      set[k] = set[k - 1];
    }
    set[i] = c;
    printPerms(set, i + 1);
    // move c back to position j
    for (int k = i; k < j; k++) {
      set[k] = set[k + 1];
    }
    set[j] = c;
  }
}