文章目录
1. 题目描述：
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
2. 思路分析：
题目的意思是实现一种正则表达式的匹配。首先对于题意：
- “a"对应"a”, 这种匹配不解释了；
- 任意字母对应".", 这也是正则常见;
- 0到多个相同字符x,对应"x*", 比起普通正则,这个地方多出来一个前缀x，x代表的是相同的字符中取一个，比如"aaaab"对应是"a*b";
- "*“还有一个易于疏忽的地方就是它的"贪婪性"要有一个限度，比如"aaa"对应"a*a”,代码逻辑不能一路贪婪到底；
正则表达式如果期望着一个字符一个字符的匹配,是非常不现实的，而"匹配"这个问题,非常容易转换成"匹配了一部分"，整个匹配不匹配,要看"剩下的匹配"情况，这就很好的把一个大的问题转换成了规模较小的问题:递归。
3. Java代码：
源代码：见我GiHub主页
代码：
public static boolean isMatch(String s, String p) {
if (p.length() == 0) {
return s.length() == 0;
}
// 表示第一个字符是否匹配上
boolean first_match = (s.length() != 0 && (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.'));
// 如果pattern是"x*"类型的话,那么pattern每次要两个两个的减少。否则,就是一个一个的减少
if (p.length() == 1 || p.charAt(1) != '*') {
return first_match && isMatch(s.substring(1), p.substring(1));
} else {
return (isMatch(s, p.substring(2)) ||
(first_match && isMatch(s.substring(1), p)));
}
}