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"
- Example:
- 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)
- Example:
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
andqueries
consist only of english letters.
这道题给了一组单词,让实现一个拼写检查器,把查询单词转换成一个正确的单词。这个拼写检查器主要有两种功能,一种是可以忽略大小写,另一种是忽略元音的错误,所谓元音是 a,e,i,o,u,这五个字母。另外题目中还制定了一些其他规则:假如有和查询单词一模一样的单词,考虑大小写,此时应该优先返回。第二个优先级是字母及顺序都一样,但大小写可能不同的,第三个优先级是有元音错误的单词也可以返回,最后都不满足的话返回空串。首先对于第一种情况,返回和查询单词一模一样的单词,很简单,将所有单词放入一个 HashSet 中,这样就可以快速确定一个查询单词是否在原单词数组中出现过。对于第二种情况,做法是将每个单词都转为小写,然后建立小写单词和原单词之间都映射,注意对于转为小写后相同都单词,我们只映射第一个出现该小写状态的单词,后面的不用管。对于第三种情况,对于每个单词,转为小写之后,然后把所有的元音字母用特殊字符替代,比如下划线,然后也是建立这种特殊处理后的状态和原单词之间的映射。当映射都建立好了之后,就可以遍历所有的查询单词了,首先是去 HashSet 中找,若有跟该查询单词一模一样的,直接加入结果 res 中。若没有,则先将查询单词变为小写,然后去第一个 HashMap 中查找,若存在,直接加入结果 res 中。若没有,再把所有的元音变为下划线,去第二个 HashMap 中查找,存在则直接加入结果 res 中。若没有,则将空串加入结果 res 中,参见代码如下:
解法一:
class Solution {
public:
vector<string> spellchecker(vector<string>& wordlist, vector<string>& queries) {
vector<string> res;
unordered_set<string> st;
unordered_map<string, string> m1;
unordered_map<string, string> m2;
for (int i = 0; i < wordlist.size(); ++i) {
string word = wordlist[i];
st.insert(word);
transform(word.begin(), word.end(), word.begin(), ::tolower);
if (!m1.count(word)) m1[word] = wordlist[i];
for (char &c : word) {
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') c = '_';
}
if (!m2.count(word)) m2[word] = wordlist[i];
}
for (string query : queries) {
if (st.count(query)) {
res.push_back(query);
continue;
}
transform(query.begin(), query.end(), query.begin(), ::tolower);
if (m1.count(query)) {
res.push_back(m1[query]);
continue;
}
for (char &c : query) {
if (c == 'a' || c == 'e' || c == 'i' || c == 'o' || c == 'u') c = '_';
}
if (m2.count(query)) {
res.push_back(m2[query]);
continue;
}
res.push_back("");
}
return res;
}
};
讨论:这里博主不得不吐槽一下 C++ 的STL的功能实在不如 Java 强大,若用 Java 写的话,可以直接调用 toLowerCase()
来转小写,同时可以用 replaceAll()
来快速替换元音,而 C++ 就显得很笨拙了,哎,谁让博主就是用 C++ 刷的顺手呢~
Github 同步地址:
https://github.com/grandyang/leetcode/issues/966
参考资料:
https://leetcode.com/problems/vowel-spellchecker/
https://leetcode.com/problems/vowel-spellchecker/discuss/211189/JavaC%2B%2BPython-Two-HashMap
LeetCode All in One 题目讲解汇总(持续更新中…)
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