标签：new ase trie tar exp opera hat reg led
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.
题意及分析：设计一个支持插入和查找的数据结构，插入的字符串只包含a-z的小写字母，查找的字符串还可包含‘.’（代表任何一个字母）。这道题可以其实就是要求一个字典树，每个树的节点有26个子节点，有一个布尔类型判断从根节点到该点组成的字符串是否是一个字母。查询时如果遇到‘.’就遍历查询当前节点的所有子节点，如果有一个符合便能查找到，否则查找不到。
代码：
class TrieNode{ public char word; //字母 public boolean isWord; //判断从根节点到该点是否是一个完整的单词 public TrieNode[] childrens = new TrieNode[26]; //子节点 public TrieNode (){} public TrieNode( char word0){ TrieNode trieNode = new TrieNode(); trieNode.word = word0; } } public class WordDictionary { TrieNode root; /** Initialize your data structure here. */ public WordDictionary() { root = new TrieNode(); root.word = ‘ ‘; } /** Adds a word into the data structure. */ public void addWord(String word) { TrieNode node = root; for(int i=0;i<word.length();){ while(i<word.length()&&node.childrens[word.charAt(i)-‘a‘]!=null){ node =node.childrens[word.charAt(i)-‘a‘]; i++; } while (i<word.length()){ node.childrens[word.charAt(i)-‘a‘] = new TrieNode(word.charAt(i)); node = node.childrens[word.charAt(i)-‘a‘]; i++; } } node.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) { TrieNode node = root; return searchRes(0,word,node); } private boolean searchRes(int start,String word,TrieNode node) { if(start == word.length()){ if(node.isWord==true) return true; else return false; } if(word.charAt(start)==‘.‘){ //如果某点为‘.‘ boolean isWord = false; for(int i=0;i<26;i++){ if(node.childrens[i]!=null){ isWord = isWord||searchRes(start+1,word,node.childrens[i]); } } return isWord; }else{ if(node.childrens[word.charAt(start)-‘a‘]!=null){ return searchRes(start+1,word,node.childrens[word.charAt(start)-‘a‘]); }else 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); */
[LeetCode] 211. Add and Search Word - Data structure design Java
标签：new ase trie tar exp opera hat reg led
原文地址：http://www.cnblogs.com/271934Liao/p/7301195.html
