How to solve the problem of regular expression matching by LeetCode

This article mainly introduces LeetCode how to solve the regular expression matching problem, has a certain reference value, interested friends can refer to, I hope you can learn a lot after reading this article, the following let the editor take you to understand it.

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Please implement a function to match the containing'. Regular expressions for 'and'. The character'.'in the pattern Represents any character, and''means that the character before it can appear any number of times (including 0 times). In this case, matching means that all characters of a string match the entire pattern. For example, the string "aaa" matches the patterns "A.A" and "abaca", but not "aa.a" and "ab*a".

Example 1: input: s = "aa" p = "a" output: false explains: "a" does not match the entire string "aa". Example 2: input: s = "aa" p = "a *" output: true explains: because'* 'represents the first element that can match zero or more, the preceding element here is' a'. Therefore, the string "aa" can be regarded as'a 'repeated once. Example 3: input: s = "ab" p = ". *" output: true explanation: ". *" means to match zero or more ('*') any character ('.'). Example 4: input: s = "aab" p = "c*a*b" output: true explanation: because'* 'means zero or more, where' c'is 0 and'a'is repeated. So you can match the string "aab". Example 5: enter: s = "mississippi" p = "mis*is*p*." Output: falses may be empty and contain only lowercase letters from Amurz. P may be empty and contain only lowercase letters and characters from Amurz. And *, there is no continuous'*'. Train of thought

If the last character of BB is a normal character, it is to see whether A [n-1] is equal to B [m-1]. If it is equal, it will look at A _ {0..n-2} and B _ {0..m-2}. If it is not equal, it will not match. This is a sub-problem.

If the last character of BB is'.', it can match any character. Look directly at A _ {0..n-2} and B _ {0..m-2}

If the last character of B is', it means that B [m-2] = c can be repeated 0 or more times, they are a whole c

Case 1: a [n-1] is 0 c 0..n-1 B the last two characters are invalid, whether it can match depends on whether A _ {0..m-3} and B _ {match} match.

Case 2: a [n-1] is the last of multiple cs (this case must be A [n-1] = c or clock.'), so A [n-1] moves forward one after the match, and B continues to match, because there can be multiple matches. Continue to see if A _ {0..n-2} and B _ {0..m-1} match.

Code class Solution {public boolean isMatch (String s, String p) {int n = s.length (), m = p.length (); boolean [] [] f = new boolean [nong1] [mCl1]; for (int I = 0; I = 2) {f [I] [j] | = f [I] [Jmur2] } if (I > = 1 & & j > = 2 & & (s.charAt (iMuth1) = = p.charAt (jmur2) | | p.charAt (jmur2) = ='.') {f [I] [j] | = f [I-1] [j] } return f [n] [m] }} Thank you for reading this article carefully. I hope the article "LeetCode how to solve the regular expression matching problem" shared by the editor will be helpful to everyone. At the same time, I also hope that you will support and follow the industry information channel. More related knowledge is waiting for you to learn!