Leet Code 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") → falseisMatch("aa","aa") → trueisMatch("aaa","aa") → falseisMatch("aa", "a*") → trueisMatch("aa", ".*") → trueisMatch("ab", ".*") → trueisMatch("aab", "c*a*b") → true

这还是个hard的问题。。。在网上找了一种中等的答案。

1. Analysis

First of all, this is one of the most difficulty problems. It is hard to think through all different cases. The problem should be simplified to handle 2 basic cases:

• the second char of pattern is "*"
• the second char of pattern is not "*"

For the 1st case, if the first char of pattern is not ".", the first char of pattern and string should be the same. Then continue to match the remaining part.

For the 2nd case, if the first char of pattern is "." or first char of pattern == the first i char of string, continue to match the remaining part.

2. Java Solution 1 (Short)

The following Java solution is accepted.

 public boolean isMatch(String s, String p) {         if(p.length() == 0)            return s.length() == 0;         //p's length 1 is special case            if(p.length() == 1 || p.charAt(1) != '*'){            if(s.length() < 1 || (p.charAt(0) != '.' && s.charAt(0) != p.charAt(0)))                return false;            return isMatch(s.substring(1), p.substring(1));             }else{            int len = s.length();             int i = -1;             while(i<len && (i < 0 || p.charAt(0) == '.' || p.charAt(0) == s.charAt(i))){                if(isMatch(s.substring(i+1), p.substring(2)))                    return true;                i++;            }            return false;        }     }

3. Java Solution 2 (More Readable)

public boolean isMatch(String s, String p) {// base caseif (p.length() == 0) {return s.length() == 0;} // special caseif (p.length() == 1) { // if the length of s is 0, return falseif (s.length() < 1) {return false;} //if the first does not match, return falseelse 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 elementif (isMatch(s, p.substring(2))) {return true;} //case 2.2: a char & '*' can stand for 1 or more preceding element, //so try every sub stringint 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;}}