缺点是，大部分的孩子阵列将大部分是空的。如果空间是一个问题时，可以使用一个更紧凑的重新presentation像一个链表，动态数组，哈希表等，但是，这些都在潜在需要若干存储器的成本和存取在每个节点处的比较，而不是以上简单的数组访问。不过，我会感到惊讶，如果浪费空间超过几兆在你的整个的字典，所以基于阵列的方法可能是你最好的选择。

有了线索，你的整个置换函数替换为一个查找，从使复杂性降低的 O（N！登录D）（其中ð是你的字典的大小， N 您的字符串的大小）为 O（N日志N）（因为你需要的字符排序，查找自身的 0 （N））。

编辑：我一起抛出的（未经测试）实现这种结构： HTTP：/ /pastebin.com/Qfu93E80

I'm writing an Android word app. My code includes a method that would find all combinations of the string and the substrings of a 7 letter string with a minimum of length 3. Then compare all available combination to every word in the dictionary to find all the valid words. I'm using a recursive method. Here's the code.

// Gets all the permutations of a string.
void permuteString(String beginningString, String endingString) {
    if (endingString.length() <= 1){
        if((Arrays.binarySearch(mDictionary, beginningString.toLowerCase() +   endingString.toLowerCase())) >= 0){
            mWordSet.add(beginningString + endingString);
        }
    }
    else
        for (int i = 0; i < endingString.length(); i++) {
            String newString = endingString.substring(0, i) + endingString.substring(i + 1);
            permuteString(beginningString + endingString.charAt(i), newString);
      }
}
// Get the combinations of the sub-strings. Minimum 3 letter combinations
void subStrings(String s){
    String newString = "";
    if(s.length() > 3){
        for(int x = 0; x < s.length(); x++){
            newString = removeCharAt(x, s);
            permuteString("", newString);
            subStrings(newString);
        }
    }
}

The above code runs fine but when I installed it on my Nexus s I realized that it runs a bit too slow. It takes a few seconds to complete. About 3 or 4 seconds which is unacceptable. Now I've played some word games on my phone and they compute all the combinations of a string instantly which makes me believe that my algorithm is not very efficient and it can be improved. Can anyone help?

public class TrieNode {
TrieNode a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z;
TrieNode[] children = {a, b, c, d, e, f, g, h, i, j, k, l, m, n, o, p, q, r, s, t, u, v, w, x, y, z};
private ArrayList<String> words = new ArrayList<String>();

public void addWord(String word){
    words.add(word);
}
public ArrayList<String> getWords(){
    return words;
}
}

public class Trie {

static String myWord;
static String myLetters = "afinnrty";
static char[] myChars;
static Sort sort;
static TrieNode myNode = new TrieNode();
static TrieNode currentNode;
static int y = 0;
static ArrayList<String> availableWords = new ArrayList<String>();

public static void main(String[] args) {

    readWords();
    getPermutations();
}
public static void getPermutations(){
    currentNode = myNode;
    for(int x = 0; x < myLetters.length(); x++){
        if(currentNode.children[myLetters.charAt(x) - 'a'] != null){
            //availableWords.addAll(currentNode.getWords());
            currentNode = currentNode.children[myLetters.charAt(x) - 'a'];
            System.out.println(currentNode.getWords() + "" + myLetters.charAt(x));
        }
    }
    //System.out.println(availableWords);
}
public static void readWords(){
    try {
        BufferedReader in = new BufferedReader(new FileReader("c://scrabbledictionary.txt"));
        String str;
        while ((str = in.readLine()) != null) {
            myWord = str;
            myChars = str.toCharArray();
            sort = new Sort(myChars);
            insert(myNode, myChars, 0);
        }
        in.close();
    } catch (IOException e) {
    }
}
public static void insert(TrieNode node, char[] myChars, int x){    
    if(x >= myChars.length){
        node.addWord(myWord);
        //System.out.println(node.getWords()+""+y);
        y++;
        return;
    }
    if(node.children[myChars[x]-'a'] == null){
        insert(node.children[myChars[x]-'a'] = new TrieNode(), myChars, x=x+1);
    }else{
        insert(node.children[myChars[x]-'a'], myChars, x=x+1);
    }
}
}

In your current approach, you're looking up every permutation of each substring. So for "abc", you need to look up "abc", "acb", "bac", "bca", "cab" and "cba". If you wanted to find all permutations of "permutations", your number of lookups is nearly 500,000,000, and that's before you've even looked at its substrings. But we can reduce this to one lookup, regardless of length, by preprocessing the dictionary.

The idea is to put each word in the dictionary into some data structure where each element contains a set of characters, and a list of all words containing (only) those characters. So for example, you could build a binary tree, which would have a node containing the (sorted) character set "abd" and the word list ["bad", "dab"]. Now, if we want to find all permutations of "dba", we sort it to give "abd" and look it up in the tree to retrieve the list.

As Baumann pointed out, tries are well suited to storing this kind of data. The beauty of the trie is that the lookup time depends only on the length of your search string - it is independent of the size of your dictionary. Since you'll be storing quite a lot of words, and most of your search strings will be tiny (the majority will be the 3-character substrings from the lowest level of your recursion), this structure is ideal.

In this case, the paths down your trie would reflect the character sets rather than the words themselves. So if your entire dictionary was ["bad", "dab", "cab", "cable"], your lookup structure would end up looking like this:

There's a bit of a time/space trade-off in the way you implement this. In the simplest (and fastest) approach, each Node contains just the list of words, and an array Node[26] of children. This allows you to locate the child you're after in constant time, just by looking at children[s.charAt(i)-'a'] (where s is your search string and i is your current depth in the trie).

The downside is that most of your children arrays will be mostly empty. If space is an issue, you can use a more compact representation like a linked list, dynamic array, hash table, etc. However, these come at the cost of potentially requiring several memory accesses and comparisons at each node, instead of the simple array access above. But I'd be surprised if the wasted space was more than a few megabytes over your whole dictionary, so the array-based approach is likely your best bet.

With the trie in place, your whole permutation function is replaced with one lookup, bringing the complexity down from O(N! log D) (where D is the size of your dictionary, N the size of your string) to O(N log N) (since you need to sort the characters; the lookup itself is O(N)).

EDIT: I've thrown together an (untested) implementation of this structure: http://pastebin.com/Qfu93E80