10. Regular Expression Matching

Implement regular expression matching with support for ‘.’ and ‘*’.

‘.’ Matches any single character.

‘*’ Matches zero or more of the preceding element.

The matching should cover the entire input string (not partial).

The function prototype should be:

bool isMatch(const char *s, const char *p)

Some examples:

isMatch(“aa”,”a”) → false

isMatch(“aa”,”aa”) → true

isMatch(“aaa”,”aa”) → false

isMatch(“aa”, “a*”) → true

isMatch(“aa”, “.*”) → true

isMatch(“ab”, “.*”) → true

isMatch(“aab”, “c*a*b”) → true

public class Solution {
public boolean isMatch(String s, String p) {
// base case
if (p.length() == 0) {
return s.length() == 0;
}

// special case
if (p.length() == 1) {

// if the length of s is 0, return false
if (s.length() < 1) {
return false;
}

// if the first does not match, return false
else if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
return false;
}

// otherwise, compare the rest of the string of s and p.
else {
return isMatch(s.substring(1), p.substring(1));
}
}

// case 1: when the second char of p is not '*'
if (p.charAt(1) != '*') {
if (s.length() < 1) {
return false;
}
if ((p.charAt(0) != s.charAt(0)) && (p.charAt(0) != '.')) {
return false;
} else {
return isMatch(s.substring(1), p.substring(1));
}
}

// case 2: when the second char of p is '*', complex case.
else {
// case 2.1: a char & '*' can stand for 0 element
if (isMatch(s, p.substring(2))) {
return true;
}

// case 2.2: a char & '*' can stand for 1 or more preceding element,
// so try every sub string
int i = 0;
while (i < s.length()
&& (s.charAt(i) == p.charAt(0) || p.charAt(0) == '.')) {
if (isMatch(s.substring(i + 1), p.substring(2))) {
return true;
}
i++;
}
return false;
}
}
}