[LeetCode] 211. Design Add and Search Words Data Structure_Medium tag: Trie, DFS

Design a data structure that supports adding new words and finding if a string matches any previously added string.

Implement the WordDictionary class:

• WordDictionary() Initializes the object.
• void addWord(word) Adds word to the data structure, it can be matched later.
• bool search(word) Returns true if there is any string in the data structure that matches word or false otherwise. word may contain dots '.' where dots can be matched with any letter.

Example:

Input
["WordDictionary","addWord","addWord","addWord","search","search","search","search"]
[[],["bad"],["dad"],["mad"],["pad"],["bad"],[".ad"],["b.."]]
Output
[null,null,null,null,false,true,true,true]

Explanation
WordDictionary wordDictionary = new WordDictionary();
wordDictionary.addWord("bad");
wordDictionary.addWord("dad");
wordDictionary.addWord("mad");
wordDictionary.search("pad"); // return False
wordDictionary.search("bad"); // return True
wordDictionary.search(".ad"); // return True
wordDictionary.search("b.."); // return True

Constraints:

• 1 <= word.length <= 500
• word in addWord consists lower-case English letters.
• word in search consist of  '.' or lower-case English letters.
• At most 50000 calls will be made to addWord and search.

Ideas:因为是word相关，所以考虑用trie来做，addWord就跟传统的Trie做的一样，不停往后加，直到最后那个node使node.isWord = True. 建立一个searchNode的helper function, 因为有‘.’， 可能需要从中间的node下去继续找， check first character, if matches, 继续找，否则return False， 如果是‘.’, 那么将所有的children都走一遍，然后recursive search (word[1:], child),如果有True，那么返回True，将所有child 都走遍没有True，最后返回False

Code:

class TrieNode:
    def __init__(self):
        self.isWord = False
        self.children = dict()
        
class WordDictionary:

    def __init__(self):
        """
        Initialize your data structure here.
        """
        self.root = TrieNode()
        

    def addWord(self, word: str) -> None:
        node = self.root
        for c in word:
            if c not in node.children:
                node.children[c] = TrieNode()
            node = node.children[c]
        node.isWord = True
        

    def search(self, word: str) -> bool:
        return self.searchNode(word, self.root)
    
    def searchNode(self, word, node) -> bool:
        if not word:
            return node.isWord
        fir = word[0]
        if fir == '.':
            for child in node.children:
                if self.searchNode(word[1:], node.children[child]):
                    return True
            return False
        elif fir in node.children:
            return self.searchNode(word[1:], node.children[fir])
        return False
            
        


# Your WordDictionary object will be instantiated and called as such:
# obj = WordDictionary()
# obj.addWord(word)
# param_2 = obj.search(word)

标签：node,Search,Medium,word,211,self,search,addWord,return
来源： https://www.cnblogs.com/Johnsonxiong/p/14952670.html