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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.
思路：刚做完前序树的题，正好拿来用，建立的时候完全一样，搜索的时候因为要对‘.‘进行判断 因此要循环所有孩子的可能，将for循环拆成每个字符判定一次的搜索。
代码：
class TrieNode{ boolean isWord; Map<Character, TrieNode> nexts; public TrieNode(){ nexts = new HashMap<Character, TrieNode>(); } } public class WordDictionary { TrieNode root = new TrieNode(); // Adds a word into the data structure. public void addWord(String word) { char[] toArray = word.toCharArray(); TrieNode temp = root; for(int i = 0 ; i < toArray.length ; i++){ if(!temp.nexts.containsKey(toArray[i])){ temp.nexts.put(toArray[i], new TrieNode()); } temp = temp.nexts.get(toArray[i]); if(i == toArray.length-1) temp.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) { return singleSearch(word, root); } public boolean singleSearch(String word, TrieNode head){ if(head==null) return false; if(word.length() == 0 ) return head.isWord; Map<Character, TrieNode> children = head.nexts; char c = word.charAt(0); if(c==‘.‘) { for(char key : children.keySet()){ //判断所有孩子的可能 if(singleSearch(word.substring(1), children.get(key) )) return true; } return false; } if(children.containsKey(c)){ //进行下一个字符的判断 return singleSearch(word.substring(1), children.get(c)); }else return false; } }
[LeetCode-JAVA] Add and Search Word - Data structure design
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原文地址：http://www.cnblogs.com/TinyBobo/p/4549427.html
