(*medium)LeetCode 211.Add and Search Word - Data structure design

Design a data structure that supports the following two operations:

void addWord(word)
bool search(word)

search(word) can search a literal word or a regular expression string containing only letters a-z or .. A . means it can represent any one letter.

For example:

addWord("bad")
addWord("dad")
addWord("mad")
search("pad") -> false
search("bad") -> true
search(".ad") -> true
search("b..") -> true

Note:
You may assume that all words are consist of lowercase letters a-z.

click to show hint.

You should be familiar with how a Trie works. If not, please work on this problem: Implement Trie (Prefix Tree) first.

方法:1：暴力解法，超时

代码如下：

public class WordDictionary {

    private List<String>list=new ArrayList<>();
    // Adds a word into the data structure.
    public void addWord(String word) {
        list.add(word);
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    public boolean search(String word) {
        if(list.contains(word)){
            return true;
        }else{
            int size=list.size();
            int i=0,j=0;
            for(;i<size;i++){
               String s= list.get(i);
               int len=s.length();
               if(len!=word.length())
                   continue;
                   j=0;
               for(;j<len;j++){
                   if(word.charAt(j)=='.')
                       continue;
                   else{
                       if(s.charAt(j)!=word.charAt(j)){
                           break;
                       }
                   }       
               }
              if(j==len) return true;
            }
            return false;
        }
    }
}

// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("pattern");

运行结果：

方法2：按照提示，使用单词查字树

代码如下：

 

public class WordDictionary {

    public class TrieNode{
        public TrieNode[] children=new TrieNode[26];
        public String item="";
    }
    private TrieNode root=new TrieNode();
    // Adds a word into the data structure.
    public void addWord(String word) {
       TrieNode node=root;
       for(char c:word.toCharArray()){
           if(node.children[c-'a']==null){
               node.children[c-'a']=new TrieNode();
           }
           node=node.children[c-'a'];
       }
       node.item=word;
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    public boolean search(String word) {
        return match(word.toCharArray(),0,root);
    }
    private boolean match(char[] chs,int k,TrieNode node){
        if(k==chs.length) return !node.item.equals("");
        if(chs[k]!='.'){
            return node.children[chs[k]-'a']!=null && match(chs,k+1,node.children[chs[k]-'a']);
        }else{
                for(int i=0;i<node.children.length;i++){
                    if(node.children[i]!=null){
                        if(match(chs,k+1,node.children[i])){
                            return true;
                        }
                    }
                }
            }
            return false;
        }
    }


// Your WordDictionary object will be instantiated and called as such:
// WordDictionary wordDictionary = new WordDictionary();
// wordDictionary.addWord("word");
// wordDictionary.search("patter");

 运行结果：