word-checker | 🇨🇳🇬🇧Chinese and English word spelling corrector | Natural Language Processing library
kandi X-RAY | word-checker Summary
kandi X-RAY | word-checker Summary
🇨🇳🇬🇧Chinese and English word spelling corrector.
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Top functions reviewed by kandi - BETA
- Delete the candidate list
- Build string
- Build String set
- Get edit result list
- Checks whether the given word is correct
- Checks if the specified text is correct
- Se string
- Static char segment
- Checks whether the word is correct
- Format a word
- Formats a character
- Gets the word collection
- Segment the given string
- Default implementation
- Correct the word
- Define English word data
- Set word format
- Correct word list
- Put candidate
- Format a character
- Init word map
- Set word data
- Compares twoDtoDto objects
- Set the word format
- Correct word
- Checks if the given word is correct
word-checker Key Features
word-checker Examples and Code Snippets
Lines of Code : 5License : Permissive (Apache-2.0)
Copyfinal String right = "正确";
final String error = "万变不离其中";
Assert.assertEquals("正确", ZhWordCheckers.correct(right));
Assert.assertEquals("万变不离其宗", ZhWordCheckers.correct(error));
final String right = "正确";
final String error = "万变不离其中";
Assert.assertTrue(ZhWordCheckers.isCorrect(right));
Assert.assertFalse(ZhWordCheckers.isCorrect(error));
com.github.houbb
word-checker
0.0.6
Community Discussions
Trending Discussions on word-checker
QUESTION
I was doing this problem on leetcode: https://leetcode.com/problems/strong-password-checker/ and I came across an edge case where "aaaaAAAAAA000000123456" was expected to take 5 steps to become a good password, according to the somewhat vague description:
deleting a character takes one step,
adding a character takes one step,
replacing a character takes one step.
The max allowed number of characters is 20 in a "strong" password, and "aaaaAAAAAA000000123456" has 22 characters. Repeating 3s like "aaa" or "OOO" are considered insecure, so to make a password secure, you'd remove, replace or add a new char to one of the characters in the repeating 3s. You also have to make sure there are capital letters, small letters, and numbers in the password to be truly secure.
According to my logic, "aaaaAAAAAA000000123456" or "aaa"+"a" +"AAA"+"AAA"+"000"+"000"+"123456" (noting the repeating 3s) should take 6 steps to become "secure", not 5:
Step 1: the string is 22 characters, so remove a char from one repeating 3 making; "aa"+"a"+"AAA"+"AAA"+"000"+"000"+"123456"
Step 2: it's now 21 chars, so remove another repeating 3 char; "aa"+"AAA"+"AAA"+"000"+"000"+"123456"
Step 3: (now 20 chars) replace a repeating 3 char; "aa"+"A1A"+"AAA"+"000"+"000"+"123456"
Step 4: replace another repeating 3 (I have to remove all of them); "aa"+"A1A"+"A2A"+"000"+"000"+"123456"
Step 5: replace another repeating 3; "aa"+"A1A"+"A2A"+"030"+"000"+"123456"
Step 6: remove the last repeating 3; "aa"+"A1A"+"A2A"+"030"+"040"+"123456"
Why should this take 5 steps and not 6?
I don't think it's necessary to add my code for this.
...ANSWER
Answered 2021-Dec-23 at 00:16aaaaAAAAAA000000123456
delete A
delete 0
aaaaAAAAA00000123456
replace a
replace A
replace 0
aazaAAzAA00z00123456
QUESTION
I am working through Leetcode problem 420. Strong Password Checker. This is the problem statement:
A password is considered strong if the below conditions are all met:
- It has at least
6characters and at most20characters.- It contains at least one lowercase letter, at least one uppercase letter, and at least one digit.
- It does not contain three repeating characters in a row (i.e.,
"...aaa..."is weak, but"...aa...a..."is strong, assuming other conditions are met).Given a string
password, return the minimum number of steps required to makepasswordstrong. ifpasswordis already strong, return0.In one step, you can:
Example 1: ...
- Insert one character to
password,- Delete one character from
password, or- Replace one character of
passwordwith another character.
ANSWER
Answered 2021-Apr-19 at 09:19It requires only one step which resolves two problems:
QUESTION
I'm am making a "word-checker" with a contenteditable div while using mark.js to highlingt certain words/word groups.
The problem I am having is that I can't find a way to "run" Mark.js as the words are being typed (copy/pasted) in.
I have tried to achieve this by adding an event listener to the textarea, but that throws an error: mark is not defined. textarea.addEventListener("input", mark); and by changing mark to instance textarea.addEventListener("input", instance); which gives me an error: TypeError: Property 'handleEvent' is not callable.. Here is the fiddle.
I have also looked at various examples found on SO and on google, but I just can't seem to find what I'm looking for. The closest thing I found was this, but I still couldn't get it to work.
Any help would be much appreciated. If I can provide any additional information, please let me know.
...ANSWER
Answered 2021-Mar-27 at 20:24You can put all the instance.mark(...) calls to a separate function (e.g. markWords).
You can then call that function when the input text changes:
Community Discussions, Code Snippets contain sources that include Stack Exchange Network
Vulnerabilities
No vulnerabilities reported
Install word-checker
You can use word-checker like any standard Java library. Please include the the jar files in your classpath. You can also use any IDE and you can run and debug the word-checker component as you would do with any other Java program. Best practice is to use a build tool that supports dependency management such as Maven or Gradle. For Maven installation, please refer maven.apache.org. For Gradle installation, please refer gradle.org .
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