[LeetCode]Add and Search Word - Data structure design,解题报告

目录

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思路

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

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思路

因为之前做了一道字典树的题目，这道题目需要借鉴字典树的数据结构。点击进入字典树参考链接。

具体思路如下：

1. 用Trie树来存储插入的字符串。
2. 用DFS来搜索Trie树中的字符串。

为什么这里只能用DFS而不能用BFS呢？

是因为：BFS无法保证下一层查找的结点是否属于当前的父节点。

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AC代码

有了字典树的基础，我直接上AC代码了，大家可以通过代码学习具体的思路：

import java.util.HashMap;class WordTrieNode {    boolean isWord;    int index;    HashMap<Character, WordTrieNode> nexts;    public WordTrieNode() {        nexts = new HashMap<Character, WordTrieNode>();    }}public class WordDictionary {    private WordTrieNode root;    public WordDictionary() {        root = new WordTrieNode();    }    // Adds a word into the data structure.    public void addWord(String word) {        WordTrieNode p = root;        int i = 0, len = word.length();        // traverse existing        while (i < len) {            char ch = word.charAt(i);            if (p.nexts.containsKey(ch)) {                p = p.nexts.get(ch);                i ++;            } else {                break;            }        }        // append new word        while (i < len) {            WordTrieNode newNode = new WordTrieNode();            newNode.index = i;            p.nexts.put(word.charAt(i), newNode);            p = newNode;            i ++;        }        // set word end        p.isWord = 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) {        if (word == null || word.length() == 0) {            return false;        }        WordTrieNode p = root;        return dfs(word, 0, p);    }       private boolean dfs(String word, int index, WordTrieNode p) {        if (index == word.length() - 1) {            if (word.charAt(index) == '.') {                for (WordTrieNode node : p.nexts.values()) {                    if (node.isWord) {                        return true;                    }                }                return false;            } else {                WordTrieNode endNode = p.nexts.get(word.charAt(index));                return endNode != null && endNode.isWord;            }        }        if (index >= word.length()) {            return false;        }        if (word.charAt(index) == '.') {            boolean res = false;            for (WordTrieNode node : p.nexts.values()) {                res |= dfs(word, index + 1, node);            }            return res;        } else {            if (p.nexts.containsKey(word.charAt(index))) {                return dfs(word, index + 1, p.nexts.get(word.charAt(index)));            } else {                return false;            }        }    }}// Your WordDictionary object will be instantiated and called as such:// WordDictionary wordDictionary = new WordDictionary();// wordDictionary.addWord("word");// wordDictionary.search("pattern");