10. Regular Expression Matching

10. Regular Expression Matching

Given an input string (s) and a pattern -, 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: falseExplanation: "a" does not match the entire string "aa".

Example 2:

Input:s = "aa"p = "a*"Output: trueExplanation: '*' means zero or more of the preceding element, 'a'. Therefore, by repeating 'a' once, it becomes "aa".

Example 3:

Input:s = "ab"p = ".*"Output: trueExplanation: ".*" means "zero or more (*) of any character (.)".

Example 4:

Input:s = "aab"p = "c*a*b"Output: trueExplanation: 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

思路： 1、“a"对应"a”, 这种匹配不解释了 2、任意字母对应".", 这也是正则常见 3、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"ab" 4、""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"aa", 代码逻辑不能一路贪婪到底 5、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 6、确定了递归以后,使用java来实现这个问题,会遇到很多和c不一样的地方,因为java对字符 的控制不像c语言指针那么灵活charAt一定要确定某个位置存在才可以使用. 7、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. 无论怎样减少,都要保证pattern有那么多个.比如s.substring(n), 其中n 最大也就是s.length()

class Solution { public boolean isMatch(String s, String p) { if (p.length() == 0) return s.length() == 0; // length == 1 is the case that is easy to forget. // as p is subtracted 2 each time, so if original // p is odd, then finally it will face the length 1 if (p.length() == 1) return (s.length() == 1) && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.'); // next char is not '*': must match current character if (p.charAt(1) != '*') { if (s.length() == 0) return false; else return (s.charAt(0) == p.charAt(0) || p.charAt(0) == '.') && isMatch(s.substring(1), p.substring(1)); }else{ // next char is * while (s.length() > 0 && (p.charAt(0) == s.charAt(0) || p.charAt(0) == '.')) { if (isMatch(s, p.substring(2))) return true; s = s.substring(1); } return isMatch(s, p.substring(2)); } }}

Java

class Solution { public boolean isMatch(String s, String p) { char[] text = s.toCharArray(); char[] pattern = p.toCharArray(); boolean T[][] = new boolean[text.length + 1][pattern.length + 1]; T[0][0] = true; // Deals with patterns like a* or a*b* or a*b*c* for (int i = 1; i < T[0].length; i++) { if (pattern[i-1] == '*') { T[0][i] = T[0][i - 2]; } } for (int i = 1; i < T.length; i++) { for (int j = 1; j < T[0].length; j++) { if (pattern[j - 1] == '.' || pattern[j - 1] == text[i - 1]) { T[i][j] = T[i - 1][j - 1]; } else if (pattern[j - 1] == '*') { T[i][j] = T[i][j - 2]; if (pattern[j-2] == '.' || pattern[j - 2] == text[i - 1]) { T[i][j] = T[i][j] | T[i - 1][j]; } } else { T[i][j] = false; } } } return T[text.length][pattern.length]; }}

JS

/** * @param {string} s * @param {string} p * @return {boolean} */var isMatch = function(s, p) { return new RegExp('^'+p+'$').test(s);};

/** * @param {string} s * @param {string} p * @return {boolean} */var isMatch = function(s, p) { const matches = new RegExp(p).exec(s); return (matches !== null && matches[0].length === s.length)};

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