10. Regular Expression Matching

Given an input string (s) and a pattern (p), 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).

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 = "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 = "c*a*b"
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 = "mis*is*p*."
Output: false

class Solution {
    public boolean isMatch(String text, String pattern) {

        //递归结束条件
        if(pattern.isEmpty()) return text.isEmpty();
        //判断 text 是否为空，防止越界，如果 text 为空， 表达式直接判为 false, text.charAt(0)就不会执行了
        boolean firstmatch = (!text.isEmpty() && (pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == '.'));
        //两种情况
        //pattern 直接跳过两个字符。表示 * 前边的字符出现 0 次
        //pattern 不变，例如 text = aa ，pattern = a*，第一个 a 匹配，然后 text 的第二个 a 接着和 pattern 的第一个 a 进行匹配。表示 * 用前一个字符替代。
        if(pattern.length() >= 2 && pattern.charAt(1) == '*'){
            return (isMatch(text, pattern.substring(2)) || firstmatch && isMatch(text.substring(1), pattern));
        }
        else    return firstmatch && isMatch(text.substring(1), pattern.substring(1));
    }
}

 

https://leetcode.windliang.cc/leetCode-10-Regular-Expression-Matching.html

class Solution {
    public boolean isMatch(String text, String pattern) {
        // 多一维的空间，因为求 dp[len - 1][j] 的时候需要知道 dp[len][j] 的情况，
        // 多一维的话，就可以把 对 dp[len - 1][j] 也写进循环了
        boolean[][] dp = new boolean[text.length() + 1][pattern.length() + 1];
        // dp[len][len] 代表两个空串是否匹配了，"" 和 "" ，当然是 true 了。
        dp[text.length()][pattern.length()] = true;

        // 从 len 开始减少
        for (int i = text.length(); i >= 0; i--) {
            for (int j = pattern.length(); j >= 0; j--) {
                // dp[text.length()][pattern.length()] 已经进行了初始化
                if(i==text.length()&&j==pattern.length()) continue;

                boolean first_match = (i < text.length() && j < pattern.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];
}

}

dp, 从后往前匹配，和从前往后匹配原则一样。

firstmatch是match第一个字符，接下来判断有✳的时候。题目要求*前面要有东西，所以非dp的情况要先判断pattern是否大于2，dp用了类似的手法