44. Wildcard Matching

问题

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)

例子

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

分析

参考10. Regular Expression Matching，其中'?'就相当于Regular Expression Matching中的'.'。'*'变成了任意字符串，包括空串。

同样可以使用动态规划：

1. 状态表
   dp[i][j]，表示s[0, i - 1]和p[0, j - 1]是否匹配

2. 初始状态
   dp[0][0] = true s为空，p为空，必然匹配；
   dp[i][0] = false, i >= 1 s非空，p为空，必然不匹配；
   dp[0][j] = p[j - 1] == '*' && dp[0][j - 1], j >= 1 s为空，只有当p[j - 1]为'*'并且dp[0][j - 1]为true时才能匹配。

3. 状态转移方程
   dp[i][j] = dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '?'); if p[j - 1] != '*'
   dp[i][j] = dp[i][j - 1] || dp[i - 1][j]; if p[j - 1] == '*' dp[i][j] = dp[i][j - 1]表示'*'被解释为空串；dp[i][j] = dp[i - 1][j]表示'*'被解释成的字符串的末尾是s[i - 1].

要点

dp

时间复杂度

O(nm)

空间复杂度

O(nm)

代码

class Solution {
public:
    bool isMatch(string s, string p) {
        int m = s.size(), n = p.size();
        vector> dp(m + 1, vector(n + 1, false));
        dp[0][0] = true;
        for (int i = 1; i <= m; i++)
            dp[i][0] = false;
        for (int j = 1; j <= n; j++)
            dp[0][j] = dp[0][j - 1] && p[j - 1] == '*';
            
        for (int i = 1; i <= m; i++) 
            for (int j = 1; j <= n; j++)
                if (p[j - 1] != '*')
                    dp[i][j] = dp[i - 1][j - 1] && (s[i - 1] == p[j - 1] || p[j - 1] == '?');
                else
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
        
        return dp[m][n];
    }
};