【刷题】Leetcode 10. Regular Expression Matching-优快云博客

本文链接：https://blog.youkuaiyun.com/qq_19450921/article/details/103229750

本文深入解析了正则表达式中'.'和'*'的匹配规则，通过多个实例展示了如何利用递归算法实现字符串与模式的精确匹配，特别强调了'*'字符在匹配零个或多个前导元素时的作用。

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‘.’ Matches any single character.
‘*’ Matches zero or more of the preceding element.
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 preceding 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) {
        if (pattern.isEmpty()) return text.isEmpty();
        boolean first_match = (!text.isEmpty() &&
                               (pattern.charAt(0) == text.charAt(0) || pattern.charAt(0) == '.'));

        if (pattern.length() >= 2 && pattern.charAt(1) == '*'){
            //这里如果*代表0个之前的符号，就是把前一个消掉，变成isMatch(text, pattern.substring(2))，比如s = "aab" p = "c*a*b"；如果代表多个之前的符号就变成first_match && isMatch(text.substring(1), pattern)，比如s = "ab" p = ".*"
            return (isMatch(text, pattern.substring(2)) ||
                    (first_match && isMatch(text.substring(1), pattern)));
        } else {
            return first_match && isMatch(text.substring(1), pattern.substring(1));
        }
    }
}