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

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<TrieNode> 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<TrieNode> 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<TrieNode>();
		}

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

		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);
	}
}

// Your WordDictionary object will be instantiated and called as such:

4 总结

很不错的问题，估计面试网易有道，360，百度这样的企业会考到。