# [966. Vowel Spellchecker](https://leetcode.com/problems/vowel-spellchecker/) ## 题目 Given a `wordlist`, we want to implement a spellchecker that converts a query word into a correct word. For a given `query` word, the spell checker handles two categories of spelling mistakes: - Capitalization: If the query matches a word in the wordlist (**case-insensitive**), then the query word is returned with the same case as the case in the wordlist. - Example: `wordlist = ["yellow"]`, `query = "YellOw"`: `correct = "yellow"` - Example: `wordlist = ["Yellow"]`, `query = "yellow"`: `correct = "Yellow"` - Example: `wordlist = ["yellow"]`, `query = "yellow"`: `correct = "yellow"` - Vowel Errors: If after replacing the vowels ('a', 'e', 'i', 'o', 'u') of the query word with any vowel individually, it matches a word in the wordlist (**case-insensitive**), then the query word is returned with the same case as the match in the wordlist. - Example: `wordlist = ["YellOw"]`, `query = "yollow"`: `correct = "YellOw"` - Example: `wordlist = ["YellOw"]`, `query = "yeellow"`: `correct = ""` (no match) - Example: `wordlist = ["YellOw"]`, `query = "yllw"`: `correct = ""` (no match) In addition, the spell checker operates under the following precedence rules: - When the query exactly matches a word in the wordlist (**case-sensitive**), you should return the same word back. - When the query matches a word up to capitlization, you should return the first such match in the wordlist. - When the query matches a word up to vowel errors, you should return the first such match in the wordlist. - If the query has no matches in the wordlist, you should return the empty string. Given some `queries`, return a list of words `answer`, where `answer[i]` is the correct word for `query = queries[i]`. **Example 1:** ``` Input:wordlist = ["KiTe","kite","hare","Hare"], queries = ["kite","Kite","KiTe","Hare","HARE","Hear","hear","keti","keet","keto"] Output:["kite","KiTe","KiTe","Hare","hare","","","KiTe","","KiTe"] ``` **Note:** - `1 <= wordlist.length <= 5000` - `1 <= queries.length <= 5000` - `1 <= wordlist[i].length <= 7` - `1 <= queries[i].length <= 7` - All strings in `wordlist` and `queries` consist only of **english** letters. ## 题目大意 在给定单词列表 wordlist 的情况下,我们希望实现一个拼写检查器,将查询单词转换为正确的单词。 对于给定的查询单词 query,拼写检查器将会处理两类拼写错误: - 大小写:如果查询匹配单词列表中的某个单词(不区分大小写),则返回的正确单词与单词列表中的大小写相同。 - 例如:wordlist = ["yellow"], query = "YellOw": correct = "yellow" - 例如:wordlist = ["Yellow"], query = "yellow": correct = "Yellow" - 例如:wordlist = ["yellow"], query = "yellow": correct = "yellow" - 元音错误:如果在将查询单词中的元音(‘a’、‘e’、‘i’、‘o’、‘u’)分别替换为任何元音后,能与单词列表中的单词匹配(不区分大小写),则返回的正确单词与单词列表中的匹配项大小写相同。 - 例如:wordlist = ["YellOw"], query = "yollow": correct = "YellOw" - 例如:wordlist = ["YellOw"], query = "yeellow": correct = "" (无匹配项) - 例如:wordlist = ["YellOw"], query = "yllw": correct = "" (无匹配项) 此外,拼写检查器还按照以下优先级规则操作: - 当查询完全匹配单词列表中的某个单词(区分大小写)时,应返回相同的单词。 - 当查询匹配到大小写问题的单词时,您应该返回单词列表中的第一个这样的匹配项。 - 当查询匹配到元音错误的单词时,您应该返回单词列表中的第一个这样的匹配项。 - 如果该查询在单词列表中没有匹配项,则应返回空字符串。 给出一些查询 queries,返回一个单词列表 answer,其中 answer[i] 是由查询 query = queries[i] 得到的正确单词。 ## 解题思路 - 读完题,很明显需要用 `map` 来解题。依题意分为 3 种情况,查询字符串完全匹配;查询字符串只是大小写不同;查询字符串有元音错误。第一种情况用 `map` `key` 直接匹配即可。第二种情况,利用 `map` 将单词从小写形式转换成原单词正确的大小写形式。第三种情况,利用 `map` 将单词从忽略元音的小写形式换成原单词正确形式。最后注意一下题目最后给的 4 个优先级规则即可。 ## 代码 ```go package leetcode import "strings" func spellchecker(wordlist []string, queries []string) []string { wordsPerfect, wordsCap, wordsVowel := map[string]bool{}, map[string]string{}, map[string]string{} for _, word := range wordlist { wordsPerfect[word] = true wordLow := strings.ToLower(word) if _, ok := wordsCap[wordLow]; !ok { wordsCap[wordLow] = word } wordLowVowel := devowel(wordLow) if _, ok := wordsVowel[wordLowVowel]; !ok { wordsVowel[wordLowVowel] = word } } res, index := make([]string, len(queries)), 0 for _, query := range queries { if _, ok := wordsPerfect[query]; ok { res[index] = query index++ continue } queryL := strings.ToLower(query) if v, ok := wordsCap[queryL]; ok { res[index] = v index++ continue } queryLV := devowel(queryL) if v, ok := wordsVowel[queryLV]; ok { res[index] = v index++ continue } res[index] = "" index++ } return res } func devowel(word string) string { runes := []rune(word) for k, c := range runes { if c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u' { runes[k] = '*' } } return string(runes) } ```