【leetcode】966. Vowel Spellchecker

标签：精确   string   color   VID   block   first   col   else   self   

题目如下：

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.

解题思路：题目给定了优先级关系，首先是精确匹配，然后是忽略大小写匹配，再接下来是忽略元音匹配。我的思路是用三个字典，第一个字典保存wordlist中所有元素，第二个字典保存wordlist把所有单词转换成小写后的新单词，第三个字典保存wordlist中的把所有单词中元音都替换成‘a‘的新单词。匹配queries中的单词也是一样的顺序，首先是精确匹配，然后忽略大小写，最后是元音都替换成‘a‘。

代码如下：

class Solution(object):
    def replaceVowel(self,v):
        newv = ‘‘
        v = v.lower()
        vowel = [‘a‘, ‘e‘, ‘i‘, ‘o‘, ‘u‘]
        for j in v:
            if j in vowel:
                newv += ‘a‘
            else:
                newv += j
        return newv
    def spellchecker(self, wordlist, queries):
        """
        :type wordlist: List[str]
        :type queries: List[str]
        :rtype: List[str]
        """
        dic = {}
        dic_case = {}
        dic_vowel = {}
        for i,v in enumerate(wordlist):
            if v not in dic:
                dic[v] = i
            if v.lower() not in dic_case:
                dic_case[v.lower()] = i
            newv = self.replaceVowel(v)
            if newv not in dic_vowel:
                dic_vowel[newv] = i
        res = []
        for i in queries:
            if i in dic:
                res.append(wordlist[dic[i]])
            elif i.lower() in dic_case:
                res.append(wordlist[dic_case[i.lower()]])
            elif self.replaceVowel(i) in dic_vowel:
                res.append(wordlist[dic_vowel[self.replaceVowel(i)]])
            else:
                res.append(‘‘)
        return res

【leetcode】966. Vowel Spellchecker

标签：精确   string   color   VID   block   first   col   else   self   

原文地址：https://www.cnblogs.com/seyjs/p/10202487.html