Simple java pattern matching (google interview)
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Simple java pattern matching (google interview)
$begingroup$
The interview question is: Write a function isMatch (without using java.util.Regex) that takes two strings as arguments: s (a string to match) and p (a regular expression pattern) and returns a boolean denoting whether s matches p. p can be composed of alphabetic characters, and '*' (denoting that the previous character matches 0 or more times).
This is what I tried, are there any bugs, or how could it be improved?
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);
}
}
java regex
asked 2 mins ago
MFXMFX
1
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Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
The interview question is: Write a function isMatch (without using java.util.Regex) that takes two strings as arguments: s (a string to match) and p (a regular expression pattern) and returns a boolean denoting whether s matches p. p can be composed of alphabetic characters, and '*' (denoting that the previous character matches 0 or more times).
This is what I tried, are there any bugs, or how could it be improved?
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);
}
}
java regex
asked 2 mins 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 |
$begingroup$
The interview question is: Write a function isMatch (without using java.util.Regex) that takes two strings as arguments: s (a string to match) and p (a regular expression pattern) and returns a boolean denoting whether s matches p. p can be composed of alphabetic characters, and '*' (denoting that the previous character matches 0 or more times).
This is what I tried, are there any bugs, or how could it be improved?
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);
}
}
java regex
asked 2 mins 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 using java.util.Regex) that takes two strings as arguments: s (a string to match) and p (a regular expression pattern) and returns a boolean denoting whether s matches p. p can be composed of alphabetic characters, and '*' (denoting that the previous character matches 0 or more times).
This is what I tried, are there any bugs, or how could it be improved?
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);
}
}
java regex
java regex
asked 2 mins 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 2 mins 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 2 mins 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 2 mins ago
MFXMFX
1
asked 2 mins 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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MFX is a new contributor. Be nice, and check out our Code of Conduct.
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