10. Regular Expression Matching
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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; } }}
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