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[LeetCode] 211. Add and Search Word - Data structure design 添加和查找单词-数据结构设计
阅读量:5152 次
发布时间:2019-06-13

本文共 4194 字,大约阅读时间需要 13 分钟。

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") -> falsesearch("bad") -> truesearch(".ad") -> truesearch("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;};

  

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转载于:https://www.cnblogs.com/lightwindy/p/9678612.html

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