leetcode面试准备:Add and Search Word – Data structure design

1 题目

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.

2 思路

该题是实现trie树的变种。因此我们首先需要定义节点的数据结构。 TrieNode节点的属性主要包含以下几点：

• char content:节点字符内容
• boolean isEnd:节点是否为一个单词结束的标记
• LinkedList childNode: 该节点的所有的孩子节点

void addWord(String word) 和Trie一样。 boolean search(String word) 采用dfs 来进行搜索。

3 代码

import java.util.LinkedList;

public class WordDictionary {
    TrieNode root;

    public WordDictionary() {
        root = new TrieNode();
    }

    // Adds a word into the data structure.
    public void addWord(String word) {
        if (search(word)) {
            return;
        }

        int len = word.length();
        TrieNode cur = root;
        for (int i = 0; i < len; i++) {
            char c = word.charAt(i);
            TrieNode tmp = cur.subNode(c);
            if (tmp == null) {
                tmp = new TrieNode(c);
                cur.childNode.add(tmp);
                cur = tmp;
            } else {
                cur = tmp;
            }
        }
        cur.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 dfs(root, word, 0);
    }

    private boolean dfs(TrieNode root, String word, int start) {
        int len = word.length();
        if (start == len) {
            if (root.isEnd)
                return true;
            else
                return false;
        }

        char c = word.charAt(start);
        if (c != '.') {
            TrieNode tmp = root.subNode(c);
            if (tmp == null)
                return false;
            else {
                return dfs(tmp, word, start + 1);
            }
        } else { // '.'
            for (TrieNode next : root.childNode) {
                boolean found = dfs(next, word, start + 1);
                if (found) {
                    return true;
                }
            }
        }
        return false;
    }

    static class TrieNode {
        // Initialize your data structure here.
        char content; // 节点内容
        boolean isEnd;// 是否为一个单词的结尾
        LinkedList childNode;// 该节点所有的孩子节点

        // Initialize your data structure here.
        public TrieNode() {
            this.content = 0;
            this.isEnd = false;
            this.childNode = new LinkedList();
        }

        public TrieNode(char content) {
            this.content = content;
            this.isEnd = false;
            this.childNode = new LinkedList();
        }

        public TrieNode subNode(char c) {
            for (TrieNode tn : childNode) {
                if (tn.content == c) {
                    return tn;
                }
            }
            return null;
        }
    }

    public static void main(String[] args) {
        WordDictionary wordDictionary = new WordDictionary();
        wordDictionary.addWord("bad");
        wordDictionary.addWord("dad");
        wordDictionary.addWord("add");
        boolean res = wordDictionary.search(".ad");
        System.out.println(res);
    }
}

4 总结

以上都是很不错的问题，估计面试网易，360，百度这样的企业会考到。