https://leetcode.com/problems/wildcard-matching/description/

狀態轉移方程考慮s[i]恰與p[j]匹配的兩種情況：相等或‘?‘，以及p[j]為‘*‘的兩種情況：s[i]為與‘*‘匹配的串的子串或‘*‘匹配一個空串即可。

主要功夫花在處理邊界上。

邊界處理格式：(?)*?*。

下午研究一下優化。

bool isMatch(string s, string p) {
    int s_length = s.size();
    int p_length = p.size();
    if (s_length == 0 && p_length == 0
 
)
        return true;
    else if (s_length != 0 && p_length == 0)
        return false;
    else if (s_length == 0 && p_length != 0){
        int flag = 0;
        for (int i = 0; i < p_length; i++)
            if (p[i] != ‘*‘)
                flag = 1;
        return flag? false: true
 
;
    }
    int status[s_length][p_length];
    memset(status, 0, (s_length)*sizeof(status[0]));
    status[0][0] = s[0] == p[0] || p[0] == ‘?‘ || p[0] == ‘*‘;
    if (status[0][0]){
        int flag = 0;
        while (++flag < p_length && p[flag] == ‘*‘)
            status[0][flag] = 1;
        if (flag < p_length && p[0
 
] == ‘*‘ && (s[0] == p[flag] || p[flag] == ‘?‘)){
            status[0][flag] = 1;
            while (++flag < p_length && p[flag] == ‘*‘)
                status[0][flag] = 1;
        }
    }
    for (int i = 1; i < s_length; i++){
        status[i][0] = p[0] == ‘*‘? 1: 0;
        for (int j = 1 ; j < p_length; j++)
            status[i][j] = (s[i] == p[j] || p[j] == ‘?‘)? status[i - 1][j - 1]: p[j] == ‘*‘? status[i][j - 1] || status[i - 1][j]: 0;
    }
    return status[s_length - 1][p_length - 1];
}

Wildcard Matching