Wildcard Matching

给出匹配字符串，使用”*“代表任意字符序列，使用”?”代表任意一个字符。给出目标字符串，判断是否匹配。
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

算法如下:

public class Solution {
    public boolean isMatch(String str, String pattern) {
        int s = 0, p = 0, match = 0, starIdx = -1;            
        while (s < str.length()){
            // advancing both pointers
            if (p < pattern.length()  && (pattern.charAt(p) == '?' || str.charAt(s) == pattern.charAt(p))){
                s++;
                p++;
            }
            // * found, only advancing pattern pointer
            else if (p < pattern.length() && pattern.charAt(p) == '*'){
                starIdx = p;
                match = s;
                p++;
            }
           // last pattern pointer was *, advancing string pointer
           // 此处一直匹配到最后一处可能一致的地方，选取的是最大范围的"*"，是贪婪的
            else if (starIdx != -1){
                p = starIdx + 1;
                match++;
                s = match;
            }
           //current pattern pointer is not star, last patter pointer was not *
          //characters do not match
            else return false;
        }
        
        //check for remaining characters in pattern
        while (p < pattern.length() && pattern.charAt(p) == '*')
            p++;
        
        return p == pattern.length(); 
    }
}