211. Add and Search Word - Data structure design

最新推荐文章于 2019-08-05 15:02:21 发布

转载最新推荐文章于 2019-08-05 15:02:21 发布 · 109 阅读

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原文链接：http://www.cnblogs.com/yuchenkit/p/7223488.html

本文介绍了一种使用Trie树的数据结构实现支持特定操作的字典：addWord用于添加单词，search用于查找指定单词或正则表达式。在遇到正则表达式中的通配符.时，会遍历当前节点的所有子节点进行匹配。

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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.

用Ｔrie树，如果碰到点（.），暴力地遍历当前ＴｒｉｅＮｏｄｅ下的所有节点

public class WordDictionary {
    class TrieNode {
        TrieNode[] children;
        int count;
        TrieNode() {
            count = 0;
            children = new TrieNode[26];
        }
    }
    TrieNode root;
    /** Initialize your data structure here. */
    public WordDictionary() {
        root = new TrieNode();
    }
    
    /** Adds a word into the data structure. */
    public void addWord(String word) {
        TrieNode cur = root;
        int i = 0;
        while (i < word.length()) {
            if (cur.children[word.charAt(i) - 'a'] == null) {
                cur.children[word.charAt(i) - 'a'] = new TrieNode();
            }
            cur = cur.children[word.charAt(i) - 'a'];
            i++;
        }
        cur.count++;
    }
    
    /** 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 searchHelper(word, 0, root);
    }
    
    public boolean searchHelper(String word, int start, TrieNode node) {
        // node != null
        int i = start;
        TrieNode cur = node;
        while (i < word.length() && word.charAt(i) != '.') {
            if (cur.children[word.charAt(i) - 'a'] == null) {
                return false;
            }
            cur = cur.children[word.charAt(i) - 'a'];
            i++;
        }
        // didn't meet '.'
        if (i == word.length()) {
            if (cur.count == 0) {
                return false;
            } else {
                return true;   
            }
        }
        // meet '.'
        for (int j = 0; j < 26; j++) {
            if (cur.children[j] != null && searchHelper(word, i + 1, cur.children[j])) {
                return true;
            }
        }
        return false;
    }
}

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

转载于:https://www.cnblogs.com/yuchenkit/p/7223488.html