leetcode:Wildcard Matching

martin_liang

于 2016-03-02 00:05:55 发布

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本文介绍了一种实现通配符匹配的算法，该算法支持 '?' 和 '*' 两种通配符，其中 '?' 可以匹配任意单个字符，而 '*' 可以匹配任意长度的字符序列。文章提供了一个 C++ 实现的示例，并通过动态规划的方法来解决这个问题。

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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)

Some examples:
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

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