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Design Add and Search Words Data Structure

Created
2022/01/30 14:26
문제 번호
211
카테고리
String
Depth-First Search
Design
Trie

Code

제출 날짜
시간
메모리
2022/01/30
104 ms
13.1 MB
// 211. Design Add and Search Words Data Structure // // https://leetcode.com/problems/design-add-and-search-words-data-structure/ // Dictionary can be easily solved with data structure map. // However, '.' replace is not that easy to be solved with just map. // '.' is just a character, trie data structure chains a character to character to make the specific word. // Such feature can ignore the '.' character, and make the comparison easier than just solving by map. // WordDictionary is the node of trie tree. // This node has several children nodes and each child node is also a node of trie tree. type WordDictionary struct { m map[byte]*WordDictionary } // Constructor function returns the root node of trie tree which is not the pointer variable. func Constructor() WordDictionary { return WordDictionary{m: make(map[byte]*WordDictionary)} } // AddWord function adds the word into the dictionary by spliting each character. // The each splitted character will be a node of trie tree. func (this *WordDictionary) AddWord(word string) { current := this for i := 0; i < len(word); i++ { w := word[i] if v, ok := current.m[w]; ok { current = v } else { current.m[w] = &WordDictionary{m: make(map[byte]*WordDictionary)} current = current.m[w] } } current.m['0'] = &WordDictionary{m: make(map[byte]*WordDictionary)} } // Search function checks whether word is exist in the dictionary or not. // This is done by recursive way. func (this *WordDictionary) Search(word string) bool { var recursion func(word string, current *WordDictionary) bool recursion = func(word string, current *WordDictionary) bool { if len(word) == 0 { return false } if word[0] == '.' { for _, v := range current.m { if v.m['0'] != nil && len(word) == 1 { return true } if recursion(word[1:], v) { return true } } } else { if v, ok := current.m[word[0]]; ok { if v.m['0'] != nil && len(word) == 1 { return true } else { return recursion(word[1:], v) } } } return false } return recursion(word, this) } // Your WordDictionary object will be instantiated and called as such: // obj := Constructor(); // obj.AddWord(word); // param_2 := obj.Search(word);
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