[LeetCode] 211. Add and Search Word - Data structure design 添加和查找单词-数据结构设计

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.

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.

208. Implement Trie (Prefix Tree) 的拓展,字典树的数据结构设计应用。

 

Java: Using backtrack to check each character of word to search.

public class WordDictionary {
    public class TrieNode {
        public TrieNode[] children = new TrieNode[26];
        public String item = "";
    }
    
    private TrieNode root = new TrieNode();

    public void addWord(String word) {
        TrieNode node = root;
        for (char c : word.toCharArray()) {
            if (node.children[c - 'a'] == null) {
                node.children[c - 'a'] = new TrieNode();
            }
            node = node.children[c - 'a'];
        }
        node.item = word;
    }

    public boolean search(String word) {
        return match(word.toCharArray(), 0, root);
    }
    
    private boolean match(char[] chs, int k, TrieNode node) {
        if (k == chs.length) return !node.item.equals("");   
        if (chs[k] != '.') {
            return node.children[chs[k] - 'a'] != null && match(chs, k + 1, node.children[chs[k] - 'a']);
        } else {
            for (int i = 0; i < node.children.length; i++) {
                if (node.children[i] != null) {
                    if (match(chs, k + 1, node.children[i])) {
                        return true;
                    }
                }
            }
        }
        return false;
    }
}  

Python:

class TrieNode:
    # Initialize your data structure here.
    def __init__(self):
        self.is_string = False
        self.leaves = {}


class WordDictionary:
    def __init__(self):
        self.root = TrieNode()

    # @param {string} word
    # @return {void}
    # Adds a word into the data structure.
    def addWord(self, word):
        curr = self.root
        for c in word:
            if c not in curr.leaves:
                curr.leaves[c] = TrieNode()
            curr = curr.leaves[c]
        curr.is_string = True

    # @param {string} word
    # @return {boolean}
    # Returns if the word is in the data structure. A word could
    # contain the dot character '.' to represent any one letter.
    def search(self, word):
        return self.searchHelper(word, 0, self.root)

    def searchHelper(self, word, start, curr):
        if start == len(word):
            return curr.is_string
        if word[start] in curr.leaves:
            return self.searchHelper(word, start+1, curr.leaves[word[start]])
        elif word[start] == '.':
            for c in curr.leaves:
                if self.searchHelper(word, start+1, curr.leaves[c]):
                    return True

        return False  

C++:

class WordDictionary {
public:
    struct TrieNode {
    public:
        TrieNode *child[26];
        bool isWord;
        TrieNode() : isWord(false) {
            for (auto &a : child) a = NULL;
        }
    };
    
    WordDictionary() {
        root = new TrieNode();
    }
    
    // Adds a word into the data structure.
    void addWord(string word) {
        TrieNode *p = root;
        for (auto &a : word) {
            int i = a - 'a';
            if (!p->child[i]) p->child[i] = new TrieNode();
            p = p->child[i];
        }
        p->isWord = true;
    }

    // Returns if the word is in the data structure. A word could
    // contain the dot character '.' to represent any one letter.
    bool search(string word) {
        return searchWord(word, root, 0);
    }
    
    bool searchWord(string &word, TrieNode *p, int i) {
        if (i == word.size()) return p->isWord;
        if (word[i] == '.') {
            for (auto &a : p->child) {
                if (a && searchWord(word, a, i + 1)) return true;
            }
            return false;
        } else {
            return p->child[word[i] - 'a'] && searchWord(word, p->child[word[i] - 'a'], i + 1);
        }
    }
    
private:
    TrieNode *root;
};

  

 类似题目:

[LeetCode] 208. Implement Trie (Prefix Tree) 实现字典树(前缀树)

 

All LeetCode Questions List 题目汇总

posted @ 2018-09-20 05:19  轻风舞动  阅读(354)  评论(0编辑  收藏  举报