leetcode【44】Wildcard Matching

问题描述：

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

题目和第十题特别像。题意就是看看s和p是否匹配。第一感觉就是动态规划的方法dp[i][j]表示s的前i个元素和p的前j个元素是否匹配。

源码：

没考虑边界条件给我整了一上午。

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