Wildcard Matching(Hard)

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", "cab") → false

解题思路1: Greedy

一个比较好的算法是贪心算法（greedy）: whenever encounter "*" in p, keep record of the current position of * in p and the current index in s(mark). Try to match the stuff behind this * in p with s, if not matched, just s++ and then try to match again

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();
}

References

1 Greedy Solution from Simple&Stupid