211. Add and Search Word - Data structure design

最新推荐文章于 2019-04-20 15:01:57 发布

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最新推荐文章于 2019-04-20 15:01:57 发布

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分类专栏： LeetCode Backtracking Design Trie 文章标签： LeetCode Backtracking Design Trie Add and Search Word

本文链接：https://blog.youkuaiyun.com/zshouyi/article/details/71915630

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本文介绍了一种数据结构WordDictionary，支持addWord和search操作。search可以使用正则表达式搜索单词，'.'代表任意一个字母。文章详细展示了如何利用Trie（前缀树）实现这一功能。

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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.

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.

首先借鉴 208. Implement Trie (Prefix Tree)的方法构建单词查找树，然后修改match函数，使之变成dfs方法，一直向下查找。遇到‘.’，遍历26个child，否则正常遍历。代码如下：

class TrieNode {
    boolean isEnd;
    TrieNode[] child;
    public TrieNode() {
        isEnd = true;
        child = new TrieNode[26];
    }
}

public class WordDictionary {
    private 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 currentNode = root;
        for (char ch: word.toCharArray()) {
            int id = ch - 'a';
            if (currentNode.child[id] == null) {
                currentNode.child[id] = new TrieNode();
                currentNode.child[id].isEnd = false;
            }
            currentNode = currentNode.child[id];
        }
        currentNode.isEnd = 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 match(word.toCharArray(), root, 0);
    }
    
    private boolean match(char[] chs, TrieNode currNode, int k) {
        if (k == chs.length) {
            return currNode.isEnd;
        }
        int id = chs[k] - 'a';
        if (chs[k] != '.') {
            if (currNode.child[id] == null) {
                return false;
            }
            return match(chs, currNode.child[id], k + 1);
        } else {
            for (int i = 0; i < 26; i ++) {
                if (currNode.child[i] != null) {
                    if (match(chs, currNode.child[i], k + 1)) {
                        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);
 */