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

通配符匹配问题，和10. Regular Expression Matching 类似。'?'匹配1个字符，'*'匹配任何字符序列，包括空的。有贪心Greedy，动态规划DP解法。

Java: Greedy, Iteration

public class WildcardMatching {
    public boolean isMatch(String s, String p) {
        int i = 0;
        int j = 0;
        int star = -1;
        int mark = -1;
        while (i < s.length()) {
            if (j < p.length()
                    && (p.charAt(j) == '?' || p.charAt(j) == s.charAt(i))) {
                ++i;
                ++j;
            } else if (j < p.length() && p.charAt(j) == '*') {
                star = j;
                j++;
                mark = i;
           //这一步是关键，匹配s中当前字符与p中‘＊’后面的字符，如果匹配，则在第一个if中处理，如果不匹配，则继续比较s中的下一个字符。
            } else if (star != -1) {
                j = star + 1;
                i = ++mark;
            } else {
                return false;
            }
        }
       //最后在此处处理多余的‘＊’，因为多个‘＊’和一个‘＊’意义相同。
        while (j < p.length() && p.charAt(j) == '*') {
            ++j;
        }
        return j == p.length();
    }
}

Python: Greedy, Iteration

class Solution:
    def isMatch(self, s, p):
        p_ptr, s_ptr, last_s_ptr, last_p_ptr = 0, 0, -1, -1
        while s_ptr < len(s):
            if p_ptr < len(p) and (s[s_ptr] == p[p_ptr] or p[p_ptr] == '?'):
                s_ptr += 1
                p_ptr += 1
            elif p_ptr < len(p) and p[p_ptr] == '*':
                p_ptr += 1
                last_s_ptr = s_ptr
                last_p_ptr = p_ptr
            elif last_p_ptr != -1:
                last_s_ptr += 1
                s_ptr = last_s_ptr
                p_ptr = last_p_ptr
            else:
                return False
            
        while p_ptr < len(p) and p[p_ptr] == '*':
            p_ptr += 1
        
        return p_ptr == len(p)

类似题目：

[LeetCode] 10. Regular Expression Matching 正则表达式匹配