# [211. Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design/)


## 题目

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`.

## 题目大意

设计一个支持以下两种操作的数据结构：`void addWord(word)`、`bool search(word)`。`search(word)` 可以搜索文字或正则表达式字符串，字符串只包含字母 . 或 a-z 。 "." 可以表示任何一个字母。



## 解题思路

- 设计一个 `WordDictionary` 的数据结构，要求具有 `addWord(word)` 和 `search(word)` 的操作，并且具有模糊查找的功能。
- 这一题是第 208 题的加强版，在第 208 题经典的 Trie 上加上了模糊查找的功能。其他实现一模一样。


## 代码

```go

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);
 */

```