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95 lines
2.5 KiB
Markdown
Executable File
95 lines
2.5 KiB
Markdown
Executable File
# [211. Add and Search Word - Data structure design](https://leetcode.com/problems/add-and-search-word-data-structure-design/)
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## 题目
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Design a data structure that supports the following two operations:
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void addWord(word)
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bool search(word)
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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.
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**Example**:
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addWord("bad")
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addWord("dad")
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addWord("mad")
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search("pad") -> false
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search("bad") -> true
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search(".ad") -> true
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search("b..") -> true
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**Note**: You may assume that all words are consist of lowercase letters `a-z`.
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## 题目大意
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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。
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## 解题思路
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- 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。
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- 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。
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## 代码
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```go
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package leetcode
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type WordDictionary struct {
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children map[rune]*WordDictionary
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isWord bool
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}
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/** Initialize your data structure here. */
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func Constructor211() WordDictionary {
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return WordDictionary{children: make(map[rune]*WordDictionary)}
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}
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/** Adds a word into the data structure. */
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func (this *WordDictionary) AddWord(word string) {
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parent := this
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for _, ch := range word {
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if child, ok := parent.children[ch]; ok {
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parent = child
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} else {
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newChild := &WordDictionary{children: make(map[rune]*WordDictionary)}
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parent.children[ch] = newChild
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parent = newChild
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}
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}
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parent.isWord = true
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}
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/** Returns if the word is in the data structure. A word could contain the dot character '.' to represent any one letter. */
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func (this *WordDictionary) Search(word string) bool {
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parent := this
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for i, ch := range word {
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if rune(ch) == '.' {
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isMatched := false
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for _, v := range parent.children {
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if v.Search(word[i+1:]) {
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isMatched = true
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}
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}
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return isMatched
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} else if _, ok := parent.children[rune(ch)]; !ok {
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return false
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}
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parent = parent.children[rune(ch)]
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}
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return len(parent.children) == 0 || parent.isWord
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}
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/**
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* Your WordDictionary object will be instantiated and called as such:
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* obj := Constructor();
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* obj.AddWord(word);
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* param_2 := obj.Search(word);
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*/
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``` |