LeetCode-44-Wildcard Matching

算法描述：

Given an input string (s) and a pattern (p), 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).

Note:

• s could be empty and contains only lowercase letters a-z.
• p could be empty and contains only lowercase letters a-z, and characters like ? or *.

Example 1:

Input:
s = "aa"
p = "a"
Output: false
Explanation: "a" does not match the entire string "aa".

Example 2:

Input:
s = "aa"
p = "*"
Output: true
Explanation: '*' matches any sequence.

Example 3:

Input:
s = "cb"
p = "?a"
Output: false
Explanation: '?' matches 'c', but the second letter is 'a', which does not match 'b'.

Example 4:

Input:
s = "adceb"
p = "*a*b"
Output: true
Explanation: The first '*' matches the empty sequence, while the second '*' matches the substring "dce".

Example 5:

Input:
s = "acdcb"
p = "a*c?b"
Output: false

解题思路：动态规划题。递推式：

dp[i][j] = dp[i-1][j-1] if(s[i-1]=p[j-1] || p[j-1]=='?')

dp[i][j] = dp[i-1][j] || dp[j][i-1] if(p[j-1]=='*')

注意初始化时，p[j-1] ='*'的处理。

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

                    
            }
        }
        return dp[m][n];
    }