Question

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.

Example:

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

Thinking:

  • Method：DFS, TrieTree
class WordDictionary {
    TrieNode root;
    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new TrieNode();
    }
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        int len = word.length();
        TrieNode temp = root;
        for(int i = 0; i < len; i++){
            int c = word.charAt(i) - 'a';
            if(temp.child[c] == null)
                temp.child[c] = new TrieNode();
            temp = temp.child[c];
        }
        temp.isLeaf = true;
    }
    /** 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 search(word, -1, root);
    }
    private boolean search(String word, int i, TrieNode node){
        if(node == null) return false;
        if(i == word.length() - 1) return node.isLeaf;
        if(word.charAt(++i) != '.'){
            return search(word, i, node.child[word.charAt(i) - 'a']);
        }else{
            for(int p = 0; p < 26; p++){
                if(search(word, i, node.child[p])){
                    if(node.child[p] == null) continue;
                    return true;
                }
            }
            return false;
        }
    }
    private static class TrieNode{
        TrieNode[] child;
        boolean isLeaf;
        TrieNode(){
            child = new TrieNode[26];
        }
    }
}
/**
 * Your WordDictionary object will be instantiated and called as such:
 * WordDictionary obj = new WordDictionary();
 * obj.addWord(word);
 * boolean param_2 = obj.search(word);
 */

二刷

  1. Use trie tree as data structure to save all the words.

  2. Since there ‘.’ might exist in the search words, we need to use dfs during searching. ```Java class WordDictionary { private static class TrieNode{ boolean isLeaf; TrieNode[] childs; public TrieNode(){ this.childs = new TrieNode[26]; } } private TrieNode root; /** Initialize your data structure here. */ public WordDictionary() { this.root = new TrieNode(); }

    /** Adds a word into the data structure. */ public void addWord(String word) { int len = word.length(); TrieNode temp = this.root; for(int i = 0; i < len; i++){ int c = word.charAt(i) - ‘a’; if(temp.childs[c] == null){ temp.childs[c] = new TrieNode(); } temp = temp.childs[c]; } temp.isLeaf = true; }

    /** 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 search(root, word, 0); }

    private boolean search(TrieNode node, String s, int index){ if(index == s.length()) return node.isLeaf; char c = s.charAt(index); if(c == ‘.’){ boolean result = false; for(int i = 0; i < 26; i++){ if(node.childs[i] != null) result |= search(node.childs[i], s, index + 1); } return result; }else{ if(node.childs[c - ‘a’] == null) return false; return search(node.childs[c - ‘a’], s, index + 1); } } }

/**

  • Your WordDictionary object will be instantiated and called as such:
  • WordDictionary obj = new WordDictionary();
  • obj.addWord(word);
  • boolean param_2 = obj.search(word); */ ```