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94 lines (85 loc) · 2.05 KB
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package algorithms;
/**
* Given a regular expression with characters a-z, ' * ', ' . '
* the task was to find if that string could match another string with characters from: a-z
* where ' * ' can delete the character before it, and '.' could match whatever character. '*' always appear after a a-z character.
* Example:
* isMatch("a*", "") = true;
* isMatch(".", "") = false;
* isMatch("ab*", "a") = true;
* isMatch("a.", "ab") = true;
* isMatch("a", "a") = true;
*
* @author joeytawadrous
*/
public class StringMatch
{
public static void main(String[] args) {
String s1 = "a*abc";
String s2 = "abc";
System.out.println(isMatch(s1, s2, 0, 0));
String s3 = "a*a*a";
String s4 = "a";
System.out.println(isMatch(s3, s4, 0, 0));
String s5 = "a.";
String s6 = "ab";
System.out.println(isMatch(s5, s6, 0, 0));
String s7 = "a*bd*c";
String s8 = "abdc";
System.out.println(isMatch(s7, s8, 0, 0));
String s9 = "a.";
String s10 = "s";
System.out.println(isMatch(s9, s10, 0, 0));
}
/**
* Checks
* @param s1
* @param s2
* @param posS1
* @param posS2
* @return boolean
*/
public static boolean isMatch(String s1, String s2, int posS1, int posS2)
{
int l1 = s1.length();
int l2 = s2.length();
// System.out.println("posS1:"+posS1+" posS2:"+posS2);
if(posS1 > l1 || posS2 > l2)
{
return false;
}
else if(posS1 == l1 && posS2 == l2)
{
return true;
}
else
{
if(posS2 < l2 && posS1 < l1 && ((s1.charAt(posS1) == s2.charAt(posS2)) || s1.charAt(posS1) == '.'))
{
return isMatch(s1, s2, posS1+1, posS2+1);
}
else if(posS1 < l1 && s1.charAt(posS1) == '*')
{
if(posS2 > 0)
{
return isMatch(s1, s2, posS1+1, posS2-1) || isMatch(s1, s2, posS1+1, posS2);
}
else
{
return isMatch(s1, s2, posS1+1, posS2);
}
}
else
{
if(posS1+1 < l1 && s1.charAt(posS1+1) == '*')
{
return isMatch(s1, s2, posS1+2, posS2);
}
else
{
return false;
}
}
}
}
}