Wildcard Matching

题目

Implement regular expression matching with support for '.' and '*'.

'.' Matches any single character.
'*' Matches zero or more of the preceding element.

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", "a*") → true
isMatch("aa", ".*") → true
isMatch("ab", ".*") → true
isMatch("aab", "c*a*b") → true

答案
几乎跟Regular Expression Matching这题是一样的，只是需要根据题目要求修改一下recurrence 的公式

class Solution {
    public boolean isMatch(String s, String p) {
        int s_len = s.length(), p_len = p.length();
        boolean[][] dp = new boolean[s_len + 1][p_len + 1];

        // Empty text matches empty empty pattern
        dp[0][0] = true;

        for(int i = 1; i <= s_len; i++)
            dp[i][0] = false;

        for(int j = 1; j <= p_len; j++) {
            char pattern_c = p.charAt(j - 1);
            if(pattern_c == '*') {
                dp[0][j] = dp[0][j - 1];
            }
            else {
                dp[0][j] = false;
            }
        }

        for(int i = 1; i <= s_len; i++) {
            for(int j = 1; j <= p_len; j++) {
                char pattern_c = p.charAt(j - 1);
                if(pattern_c == '*') {
                    dp[i][j] = dp[i][j - 1] || dp[i - 1][j];
                }
                else if(pattern_c == '?' || pattern_c == s.charAt(i - 1)) {
                    dp[i][j] = dp[i - 1][j - 1];
                }
            }
        }

        return dp[s_len][p_len];
    }
}