0211. Design Add and Search Words Data Structure

211. Design Add and Search Words Data Structure#

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")
search("b..") -> true
``````

Note: You may assume that all words are consist of lowercase letters `a-z`.

解题思路 #

• 设计一个 `WordDictionary` 的数据结构，要求具有 `addWord(word)``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 #

``````
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();