Simple pattern matching between two string inputs using Java (Google interview challenge)
0
$begingroup$
The interview question is:
Write a function
isMatch(without usingjava.util.Regex) that takes two strings as arguments:
s(a string to match), and
p(a regular expression pattern).
It should return a Boolean denoting whether
smatchesp.
pcan be composed of alphabetic characters and '*' (denoting that the previous character matches 0 or more times).
This is the code I have written:
private boolean isNextCharStarred(String s) {
return s.length() >= 2 && s.charAt(1) == '*';
}
private boolean recIsMatch(String s, String p, boolean hasPatternBeenEaten) {
if (p.length() == 0) {
return true;
}
if (s.length() == 0) {
if (isNextCharStarred(p)) {
return recIsMatch(s, p.substring(2), true);
}
return p.length() == 0;
}
if (isNextCharStarred(p)) {
if (s.charAt(0) != p.charAt(0)) {
return recIsMatch(s, p.substring(2), true);
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten)
|| recIsMatch(s, p.substring(2), true);
}
if (s.charAt(0) == p.charAt(0)) {
return recIsMatch(s.substring(1), p.substring(1), true);
}
if (hasPatternBeenEaten) {
return false;
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten);
}
public boolean isMatch(String s, String p) {
return recIsMatch(s, p, false);
}
Are there any bugs? How could it be improved?
java strings interview-questions reinventing-the-wheel
edited 15 mins ago
Cody Gray
3,433926
asked 4 hours ago
MFXMFX
1
New contributor
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add a comment |
0
$begingroup$
The interview question is:
Write a function
isMatch(without usingjava.util.Regex) that takes two strings as arguments:
s(a string to match), and
p(a regular expression pattern).
It should return a Boolean denoting whether
smatchesp.
pcan be composed of alphabetic characters and '*' (denoting that the previous character matches 0 or more times).
This is the code I have written:
private boolean isNextCharStarred(String s) {
return s.length() >= 2 && s.charAt(1) == '*';
}
private boolean recIsMatch(String s, String p, boolean hasPatternBeenEaten) {
if (p.length() == 0) {
return true;
}
if (s.length() == 0) {
if (isNextCharStarred(p)) {
return recIsMatch(s, p.substring(2), true);
}
return p.length() == 0;
}
if (isNextCharStarred(p)) {
if (s.charAt(0) != p.charAt(0)) {
return recIsMatch(s, p.substring(2), true);
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten)
|| recIsMatch(s, p.substring(2), true);
}
if (s.charAt(0) == p.charAt(0)) {
return recIsMatch(s.substring(1), p.substring(1), true);
}
if (hasPatternBeenEaten) {
return false;
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten);
}
public boolean isMatch(String s, String p) {
return recIsMatch(s, p, false);
}
Are there any bugs? How could it be improved?
java strings interview-questions reinventing-the-wheel
edited 15 mins ago
Cody Gray
3,433926
asked 4 hours ago
MFXMFX
1
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
0
0
0
$begingroup$
The interview question is:
Write a function
isMatch(without usingjava.util.Regex) that takes two strings as arguments:
s(a string to match), and
p(a regular expression pattern).
It should return a Boolean denoting whether
smatchesp.
pcan be composed of alphabetic characters and '*' (denoting that the previous character matches 0 or more times).
This is the code I have written:
private boolean isNextCharStarred(String s) {
return s.length() >= 2 && s.charAt(1) == '*';
}
private boolean recIsMatch(String s, String p, boolean hasPatternBeenEaten) {
if (p.length() == 0) {
return true;
}
if (s.length() == 0) {
if (isNextCharStarred(p)) {
return recIsMatch(s, p.substring(2), true);
}
return p.length() == 0;
}
if (isNextCharStarred(p)) {
if (s.charAt(0) != p.charAt(0)) {
return recIsMatch(s, p.substring(2), true);
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten)
|| recIsMatch(s, p.substring(2), true);
}
if (s.charAt(0) == p.charAt(0)) {
return recIsMatch(s.substring(1), p.substring(1), true);
}
if (hasPatternBeenEaten) {
return false;
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten);
}
public boolean isMatch(String s, String p) {
return recIsMatch(s, p, false);
}
Are there any bugs? How could it be improved?
java strings interview-questions reinventing-the-wheel
edited 15 mins ago
Cody Gray
3,433926
asked 4 hours ago
MFXMFX
1
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
The interview question is:
Write a function
isMatch(without usingjava.util.Regex) that takes two strings as arguments:
s(a string to match), and
p(a regular expression pattern).
It should return a Boolean denoting whether
smatchesp.
pcan be composed of alphabetic characters and '*' (denoting that the previous character matches 0 or more times).
This is the code I have written:
private boolean isNextCharStarred(String s) {
return s.length() >= 2 && s.charAt(1) == '*';
}
private boolean recIsMatch(String s, String p, boolean hasPatternBeenEaten) {
if (p.length() == 0) {
return true;
}
if (s.length() == 0) {
if (isNextCharStarred(p)) {
return recIsMatch(s, p.substring(2), true);
}
return p.length() == 0;
}
if (isNextCharStarred(p)) {
if (s.charAt(0) != p.charAt(0)) {
return recIsMatch(s, p.substring(2), true);
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten)
|| recIsMatch(s, p.substring(2), true);
}
if (s.charAt(0) == p.charAt(0)) {
return recIsMatch(s.substring(1), p.substring(1), true);
}
if (hasPatternBeenEaten) {
return false;
}
return recIsMatch(s.substring(1), p, hasPatternBeenEaten);
}
public boolean isMatch(String s, String p) {
return recIsMatch(s, p, false);
}
Are there any bugs? How could it be improved?
java strings interview-questions reinventing-the-wheel
java strings interview-questions reinventing-the-wheel
edited 15 mins ago
Cody Gray
3,433926
asked 4 hours ago
MFXMFX
1
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 15 mins ago
Cody Gray
3,433926
asked 4 hours ago
MFXMFX
1
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 15 mins ago
Cody Gray
3,433926
edited 15 mins ago
Cody Gray
3,433926
edited 15 mins ago
Cody Gray
3,433926
3,433926
asked 4 hours ago
MFXMFX
1
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 4 hours ago
MFXMFX
1
asked 4 hours ago
MFXMFX
1
1
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
MFX is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
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