44. Wildcard Matching

problem description

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

algorithm thought

这题很之前的正则表达式很像，其实还简单一点。使用二维数组保存中间结果，动态规划解决问题。可以借鉴前面动态规划题目的解答。这里的?其实和之前的.一样。这里的*比之前的*更加加单

code

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

algorithm analysis

时间复杂度O(n²)