211. Design Add and Search Words Data Structure

Problem

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.

Problem Summary

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; the string contains only the letter . or a-z. "." can represent any one letter.

Solution Approach

Design a WordDictionary data structure, requiring the operations addWord(word) and search(word), and also supporting fuzzy search.
This problem is an enhanced version of Problem 208, adding fuzzy search functionality to the classic Trie from Problem 208. The rest of the implementation is exactly the same.

Code


package leetcode

type WordDictionary struct {
	children map[rune]*WordDictionary
	isWord   bool
}

/** Initialize your data structure here. */
func Constructor211() WordDictionary {
	return WordDictionary{children: make(map[rune]*WordDictionary)}
}

/** Adds a word into the data structure. */
func (this *WordDictionary) AddWord(word string) {
	parent := this
	for _, ch := range word {
		if child, ok := parent.children[ch]; ok {
			parent = child
		} else {
			newChild := &WordDictionary{children: make(map[rune]*WordDictionary)}
			parent.children[ch] = newChild
			parent = newChild
		}
	}
	parent.isWord = true
}

/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
func (this *WordDictionary) Search(word string) bool {
	parent := this
	for i, ch := range word {
		if rune(ch) == '.' {
			isMatched := false
			for _, v := range parent.children {
				if v.Search(word[i+1:]) {
					isMatched = true
				}
			}
			return isMatched
		} else if _, ok := parent.children[rune(ch)]; !ok {
			return false
		}
		parent = parent.children[rune(ch)]
	}
	return len(parent.children) == 0 || parent.isWord
}

/**
 * Your WordDictionary object will be instantiated and called as such:
 * obj := Constructor();
 * obj.AddWord(word);
 * param_2 := obj.Search(word);
 */