Determine whether a word appears in a sentence
Given an array of words representing a sentence and a word to search for, the method must determine whether the word appears in the sentence.
Specifications
- Case insensitive.
- The arguments are never
null. - The sentence is in the English language.
/**
* A case-insensitive search on a sentence to find whether a given word can be found in it. It
* keeps certain punctuation in mind.
*
* @param aSentence An array of words forming a sentence. Cannot be {@code null}.
* @param aWord The word to search for. Cannot be {@code null}.
*
* @return {@code true} if the word appeared at least once in the sentence, {@code false}
* otherwise.
*/
public boolean sentenceContainsWord(final String aSentence, String aWord)
{
// The search is case-insensitive
aWord = aWord.toLowerCase();
// Loop through the words in the sentence
for (int i = 0; i < aSentence.length; i++)
{
String sentenceWord = aSentence[i];
// If the word in the sentence matches, return immediately
if (sentenceWord.toLowerCase().equals(aWord))
{
return true;
}
// The word could end with punctuation like a comma or a dot
int lastSentenceWordIndex = sentenceWord.length() - 1;
String lastCharacter = Character.toString(sentenceWord.charAt(lastSentenceWordIndex));
if (lastCharacter.matches("[.,:;]") && sentenceWord.substring(0, lastSentenceWordIndex).equals(aWord))
{
return true;
}
}
return false;
}
java
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Given an array of words representing a sentence and a word to search for, the method must determine whether the word appears in the sentence.
Specifications
- Case insensitive.
- The arguments are never
null. - The sentence is in the English language.
/**
* A case-insensitive search on a sentence to find whether a given word can be found in it. It
* keeps certain punctuation in mind.
*
* @param aSentence An array of words forming a sentence. Cannot be {@code null}.
* @param aWord The word to search for. Cannot be {@code null}.
*
* @return {@code true} if the word appeared at least once in the sentence, {@code false}
* otherwise.
*/
public boolean sentenceContainsWord(final String aSentence, String aWord)
{
// The search is case-insensitive
aWord = aWord.toLowerCase();
// Loop through the words in the sentence
for (int i = 0; i < aSentence.length; i++)
{
String sentenceWord = aSentence[i];
// If the word in the sentence matches, return immediately
if (sentenceWord.toLowerCase().equals(aWord))
{
return true;
}
// The word could end with punctuation like a comma or a dot
int lastSentenceWordIndex = sentenceWord.length() - 1;
String lastCharacter = Character.toString(sentenceWord.charAt(lastSentenceWordIndex));
if (lastCharacter.matches("[.,:;]") && sentenceWord.substring(0, lastSentenceWordIndex).equals(aWord))
{
return true;
}
}
return false;
}
java
New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What about punctuation inside or in front of a word?'Like,' he said 'RobAu's comment?'?
– RobAu
yesterday
Also, if to pass an empty-string as word, you will get anIndexOutOfBoundsException
– RobAu
yesterday
add a comment |
Given an array of words representing a sentence and a word to search for, the method must determine whether the word appears in the sentence.
Specifications
- Case insensitive.
- The arguments are never
null. - The sentence is in the English language.
/**
* A case-insensitive search on a sentence to find whether a given word can be found in it. It
* keeps certain punctuation in mind.
*
* @param aSentence An array of words forming a sentence. Cannot be {@code null}.
* @param aWord The word to search for. Cannot be {@code null}.
*
* @return {@code true} if the word appeared at least once in the sentence, {@code false}
* otherwise.
*/
public boolean sentenceContainsWord(final String aSentence, String aWord)
{
// The search is case-insensitive
aWord = aWord.toLowerCase();
// Loop through the words in the sentence
for (int i = 0; i < aSentence.length; i++)
{
String sentenceWord = aSentence[i];
// If the word in the sentence matches, return immediately
if (sentenceWord.toLowerCase().equals(aWord))
{
return true;
}
// The word could end with punctuation like a comma or a dot
int lastSentenceWordIndex = sentenceWord.length() - 1;
String lastCharacter = Character.toString(sentenceWord.charAt(lastSentenceWordIndex));
if (lastCharacter.matches("[.,:;]") && sentenceWord.substring(0, lastSentenceWordIndex).equals(aWord))
{
return true;
}
}
return false;
}
java
New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Given an array of words representing a sentence and a word to search for, the method must determine whether the word appears in the sentence.
Specifications
- Case insensitive.
- The arguments are never
null. - The sentence is in the English language.
/**
* A case-insensitive search on a sentence to find whether a given word can be found in it. It
* keeps certain punctuation in mind.
*
* @param aSentence An array of words forming a sentence. Cannot be {@code null}.
* @param aWord The word to search for. Cannot be {@code null}.
*
* @return {@code true} if the word appeared at least once in the sentence, {@code false}
* otherwise.
*/
public boolean sentenceContainsWord(final String aSentence, String aWord)
{
// The search is case-insensitive
aWord = aWord.toLowerCase();
// Loop through the words in the sentence
for (int i = 0; i < aSentence.length; i++)
{
String sentenceWord = aSentence[i];
// If the word in the sentence matches, return immediately
if (sentenceWord.toLowerCase().equals(aWord))
{
return true;
}
// The word could end with punctuation like a comma or a dot
int lastSentenceWordIndex = sentenceWord.length() - 1;
String lastCharacter = Character.toString(sentenceWord.charAt(lastSentenceWordIndex));
if (lastCharacter.matches("[.,:;]") && sentenceWord.substring(0, lastSentenceWordIndex).equals(aWord))
{
return true;
}
}
return false;
}
java
java
New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Skere is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
What about punctuation inside or in front of a word?'Like,' he said 'RobAu's comment?'?
– RobAu
yesterday
Also, if to pass an empty-string as word, you will get anIndexOutOfBoundsException
– RobAu
yesterday
add a comment |
What about punctuation inside or in front of a word?'Like,' he said 'RobAu's comment?'?
– RobAu
yesterday
Also, if to pass an empty-string as word, you will get anIndexOutOfBoundsException
– RobAu
yesterday
What about punctuation inside or in front of a word?
'Like,' he said 'RobAu's comment?' ?– RobAu
yesterday
What about punctuation inside or in front of a word?
'Like,' he said 'RobAu's comment?' ?– RobAu
yesterday
Also, if to pass an empty-string as word, you will get an
IndexOutOfBoundsException– RobAu
yesterday
Also, if to pass an empty-string as word, you will get an
IndexOutOfBoundsException– RobAu
yesterday
add a comment |
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What about punctuation inside or in front of a word?
'Like,' he said 'RobAu's comment?'?– RobAu
yesterday
Also, if to pass an empty-string as word, you will get an
IndexOutOfBoundsException– RobAu
yesterday