标签:
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
解题思路:
参考之前的Java for LeetCode 208 Implement Trie (Prefix Tree) 修改下即可,JAVA实现如下:
public class WordDictionary extends Trie { public void addWord(String word) { super.insert(word); } // 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; return search(word, 0, root); } public boolean search(String word, int depth, TrieNode node) { if (depth == word.length() - 1) { if (word.charAt(depth) != ‘.‘) { if (node.son[word.charAt(depth) - ‘a‘] != null) { node = node.son[word.charAt(depth) - ‘a‘]; return node.isEnd; } else return false; } for (int i = 0; i < 26; i++) { if (node.son[i] != null) { TrieNode ason = node.son[i]; if (ason.isEnd) return true; } } return false; } if (word.charAt(depth) != ‘.‘) { if (node.son[word.charAt(depth) - ‘a‘] != null) { node = node.son[word.charAt(depth) - ‘a‘]; return search(word, depth + 1, node); } else return false; } for (int i = 0; i < 26; i++) { if (node.son[i] != null) { TrieNode ason = node.son[i]; if (search(word, depth + 1, ason)) return true; } } return false; } } class TrieNode { // Initialize your data structure here. int num;// 有多少单词通过这个节点,即节点字符出现的次数 TrieNode[] son;// 所有的儿子节点 boolean isEnd;// 是不是最后一个节点 char val;// 节点的值 TrieNode() { this.num = 1; this.son = new TrieNode[26]; this.isEnd = false; } } class Trie { protected TrieNode root; public Trie() { root = new TrieNode(); } public void insert(String word) { if (word == null || word.length() == 0) return; TrieNode node = this.root; char[] letters = word.toCharArray(); for (int i = 0; i < word.length(); i++) { int pos = letters[i] - ‘a‘; if (node.son[pos] == null) { node.son[pos] = new TrieNode(); node.son[pos].val = letters[i]; } else { node.son[pos].num++; } node = node.son[pos]; } node.isEnd = true; } // Returns if the word is in the trie. public boolean search(String word) { if (word == null || word.length() == 0) { return false; } TrieNode node = root; char[] letters = word.toCharArray(); for (int i = 0; i < word.length(); i++) { int pos = letters[i] - ‘a‘; if (node.son[pos] != null) { node = node.son[pos]; } else { return false; } } return node.isEnd; } // Returns if there is any word in the trie // that starts with the given prefix. public boolean startsWith(String prefix) { if (prefix == null || prefix.length() == 0) { return false; } TrieNode node = root; char[] letters = prefix.toCharArray(); for (int i = 0; i < prefix.length(); i++) { int pos = letters[i] - ‘a‘; if (node.son[pos] != null) { node = node.son[pos]; } else { return false; } } return true; } } // Your Trie object will be instantiated and called as such: // Trie trie = new Trie(); // trie.insert("somestring"); // trie.search("key");
Java for LeetCode 211 Add and Search Word - Data structure design
标签:
原文地址:http://www.cnblogs.com/tonyluis/p/4564310.html