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 8 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 8 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 8 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 8 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 8 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 8 mins ago
Cody Gray
3,433926
edited 8 mins ago
Cody Gray
3,433926
edited 8 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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