leetcode题解（六）： Regular Expression Matching

题目描述：正则表达式匹配
难度：hard
例子：
Example 1:

Input:
s = “aa”
p = “a”
Output: false
Explanation: “a” does not match the entire string “aa”.
Example 2:

Input:
s = “aa”
p = “a*”
Output: true
Explanation: ‘*’ means zero or more of the precedeng element, ‘a’. Therefore, by repeating ‘a’ once, it becomes “aa”.
Example 3:

Input:
s = “ab”
p = “."
Output: true
Explanation: ".” means “zero or more (*) of any character (.)”.
Example 4:

Input:
s = “aab”
p = “cab”
Output: true
Explanation: c can be repeated 0 times, a can be repeated 1 time. Therefore it matches “aab”.
Example 5:

Input:
s = “mississippi”
p = “misisp*.”
Output: false

class Solution {
    public boolean isMatch(String text, String pattern) {
        boolean[][] dp = new boolean[text.length() + 1][pattern.length() + 1];
        dp[text.length()][pattern.length()] = true;

        for (int i = text.length(); i >= 0; i--){
            for (int j = pattern.length() - 1; j >= 0; j--){
                boolean first_match = (i < text.length() &&
                                       (pattern.charAt(j) == text.charAt(i) ||
                                        pattern.charAt(j) == '.'));
                if (j + 1 < pattern.length() && pattern.charAt(j+1) == '*'){
                    dp[i][j] = dp[i][j+2] || first_match && dp[i+1][j];
                } else {
                    dp[i][j] = first_match && dp[i+1][j+1];
                }
            }
        }
        return dp[0][0];
    }
}