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

題解
看題目不難想到用dp的方法，需要想到特殊樣例。例如

Input:
s = "”
p = "**"
Output: true

就是有連續多個*在開頭，需要做個預處理操作，否則過不了。

dp陣列構造
dp[i][j] 代表p從[1…i]，s從[1…j]，這兩個字串是否匹配（字串下標從1開始）
dp[0][0] 代表兩個空字串，顯然是匹配的。
dp[0][i] 代表p串為空
dp[i][0] 代表s串為空

if s[j] == p[i] || p[i] == '?'
	dp[i][j] = dp[i-1][j-1]  // 去掉s，p末端
else if p[i] == '*'
	dp[i][j] = 1 if dp[i-1][k] == 1 （0 <= k <= j） // 去掉p末端

程式碼

class Solution {
    bool dp[2000][2000]={0};
public:
    bool isMatch(string s1, string s2) {
        int n=s1.size(), m=s2.size();
        for(int j=0;s2[j]=='*' && j<m;j++) // 開頭預處理，重要
            for(int i=0;i<=n;i++)
                dp[j+1][i]=true;
        dp[0][0] = true;
        for(int i=0;i<m;i++)
            for(int j=0;j<n;j++) 
                if(s2[i]=='*') 
                    for(int k=0;!dp[i+1][j+1]&&k<=j+1;k++) 
                        dp[i+1][j+1]=dp[i][k];          
                else if(s2[i]=='?'||s1[j]==s2[i]) 
                    dp[i+1][j+1]=dp[i][j];                     
        return dp[m][n];
    }
};