2

在不重复字母的情况下尝试显示数组字符的每个组合时遇到问题。

public static String[] getAllLists(String[] elements, int lengthOfList)
{
    //initialize our returned list with the number of elements calculated above
    String[] allLists = new String[(int)Math.pow(elements.length, lengthOfList)];

    //lists of length 1 are just the original elements
    if(lengthOfList == 1) return elements; 
    else
    {
        //the recursion--get all lists of length 3, length 2, all the way up to 1
        String[] allSublists = getAllLists(elements, lengthOfList - 1);

        //append the sublists to each element
        int arrayIndex = 0;

        for(int i = 0; i < elements.length; i++)
        {
            for(int j = 0; j < allSublists.length; j++)
            {
                //add the newly appended combination to the list
                allLists[arrayIndex] = elements[i] + allSublists[j];
                arrayIndex++;
            }
        }

        return allLists;
    }
}

上面的代码可以完美运行，但是在这种情况下不能多次使用每个字母。

我现在被困住了如何做到这一点。

4

1 回答 1

10

这是一个示例实现。本质上，它需要一个字符串并遍历每个字符，将该字符放在前面。然后它在剩余的字符上递归。该结构消除了重复字母的问题，因为递归的输入已经删除了您已经使用的字符。

我还将结果存储在一个集合中以删除语义等价。输入 'aab' 可以切换 char 0 和 char 1 但仍然是 'aab'。我使用 TreeSet 来保留排序以便更轻松地验证输出，但 HashSet 会更快。

  public static Set<String> permute(String chars)
  {
    // Use sets to eliminate semantic duplicates (aab is still aab even if you switch the two 'a's)
    // Switch to HashSet for better performance
    Set<String> set = new TreeSet<String>();

    // Termination condition: only 1 permutation for a string of length 1
    if (chars.length() == 1)
    {
      set.add(chars);
    }
    else
    {
      // Give each character a chance to be the first in the permuted string
      for (int i=0; i<chars.length(); i++)
      {
        // Remove the character at index i from the string
        String pre = chars.substring(0, i);
        String post = chars.substring(i+1);
        String remaining = pre+post;

        // Recurse to find all the permutations of the remaining chars
        for (String permutation : permute(remaining))
        {
          // Concatenate the first character with the permutations of the remaining chars
          set.add(chars.charAt(i) + permutation);
        }
      }
    }
    return set;
  }

示例运行：

  public static void main(String[] args)
  {
    for (String s : CharPermuter.permute("abca"))
    {
      System.out.println(s);
    }
  }

生成：

aabc
aacb
abac
abca
acab
acba
baac
baca
bcaa
caab
caba
cbaa
于 2012-03-15T17:09:22.247 回答