leetcode--Wildcard Matching

Implement wildcard pattern matching with support for '?' and '*'.

'?' Matches any single character.
'*' Matches any sequence of characters (including the empty sequence).

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", "*") → true
isMatch("aa", "a*") → true
isMatch("ab", "?*") → true
isMatch("aab", "c*a*b") → false

dynamic programming code

public class Solution {
    public boolean isMatch(String s, String p) {
		//check the characters in p which is not *
		int count = 0;
		for(int i = 0; i < p.length(); ++i)
			count += (p.charAt(i) == '*') ? 0 : 1;
		
		if(count > s.length())
			return false;
		
		boolean[][] matchChecked = new boolean[s.length() + 1][p.length() + 1];
		matchChecked[0][0] = true;
		for(int i = 1; i < s.length() + 1; ++i)
			matchChecked[i][0] = false;
		for(int i = 1; i < p.length() + 1; ++i) 
			matchChecked[0][i]  = matchChecked[0][i - 1] && (p.charAt(i - 1) == '*');
		for(int i = 1; i < s.length() + 1; ++i) {
			for(int j = 1; j < p.length() + 1; ++j) {
				if(p.charAt(j - 1) == '*')
					matchChecked[i][j] = matchChecked[i - 1][j - 1] || matchChecked[i - 1][j] || matchChecked[i][j - 1];
				else
					matchChecked[i][j] = (matchChecked[i - 1][j - 1] && (p.charAt(j - 1) == '?'))
									  || (matchChecked[i - 1][j - 1] && (p.charAt(j - 1) == s.charAt(i - 1)));
			}
		}
		return matchChecked[s.length()][p.length()];
    }
}

 

another code:

public class Solution {
    public boolean isMatch(String s, String p) {
        int slen = s.length(), plen = p.length();
        int start = Integer.MAX_VALUE, resume = 0; 
        int i = 0, j = 0;
        while(i < slen){
            //remove the *'s in roll, keep the index of i for resume the further outer loop
            while(j< plen && p.charAt(j) == '*'){
                start = j;
                ++j;
                resume = i;
            }
            
            //when j reach the end of the string p and             
            if(j == plen ||(p.charAt(j) != s.charAt(i) && p.charAt(j) != '?')){
                //if there is no * in p, then these two strings are not matched under the above 
                //condition
                if(start > plen)
                    return false;
                //if there is a * in p, we need ignore some characters in s and resume comparison.
                else{
                    j = start + 1;
                    ++resume; //This is important, we need to resume the comparsion from each 
                              //possible position after i
                    i = resume;
                }    
            }
            //when two strings are same at theses positions. move i and j forward simultanenously.
            else{
                ++i;
                ++j;
            }
        }
        
        //Once we exhausted the string s, we need to check whether  p reaches its end
        while(j < plen && p.charAt(j) == '*')
            ++j;
        return j == plen;
    }
}