Interview 10 Big algorithm Rollup-strings and Arrays 3

6. Matching Regular expressions

Write the function Boolismatch (string s, String p) to complete the regular expression "." And the function of "*". Cases:

IsMatch ("AA", "a") return false

IsMatch ("AA", "AA") return True

IsMatch ("AAA", "AA") return False

IsMatch ("AA", "A *") return True

IsMatch ("AA", ". *") return True

IsMatch ("AB", ". *") return True

IsMatch ("AaB", "C*a*b") return True

Note: The first character of P is not *, and there is no two * connected in P

The problem can be divided into the following scenarios:

1. The length of P is 0: at this time the length of s must be 0 to match

2. The length of P is 1: If the length of S is less than 1, it must not match. Otherwise, if P.charat (0) is not "." Or not equal to S is not matched

3. Length of P >1:

Here are two possible considerations: I.p.charat (1)! = ' * '

Ii.p.charat (1) = = ' *

When P.charat (1)! = ' * ', the length of S is less than 1, it is bound to be mismatched. Compares the first character of S with P: if S is different from the first character of P and the first character of P is not ".", then it does not match. Otherwise iterate calls the function, return IsMatch (s.substring (1), p.substring (1))

When P.charat (1) = = ' * ',

。。。

PS: There is a problem with the code:

http://www.programcreek.com/2012/12/leetcode-regular-expression-matching-in-java/

For example, the following code test failed, temporarily did not think of the solution:

public class Test {public static Boolean IsMatch (string s, String p) {//Base Caseif (p.length () = = 0) {return s.length () = = 0;} Special Caseif (p.length () = = 1) {//If the length of S is 0, return Falseif (S.length () < 1) {return false;}//if The first does not match, return Falseelse if ((P.charat (0)! = S.charat (0)) && (P.charat (0)! = '. ')) {return false;} Otherwise, compare the rest of the string of S and P.else {return IsMatch (s.substring (1), p.substring (1));}} Case 1:when the second char of P was not ' * ' if (P.charat (1)! = ' * ') {if (S.length () < 1) {return false;} if ((P.charat (0)! = S.charat (0)) && (P.charat (0)! = '. ')) {return false;} else {return IsMatch (s.substring (1), p.substring (1));}} Case 2:when the second char of P was ' * ', complex case.else {//case 2.1:a char & ' * ' can stand for 0 elementif (is Match (S, p.substring (2))) {return true;}//case 2.2:a char & ' * ' can stand for 1 or more preceding element,//so try Every sub Stringinti = 0;while (I<s.length () && (S.charat (i) ==p.charat (0) | | P.charat (0) = = '. ')) {if (IsMatch (s.substring (i + 1), p.substring (2))) {return true;} i++;} return false;}} public static void Main (string[] args) {System.out.println (IsMatch ("12", "1*12"));}}

Interview 10 Big algorithm Rollup-strings and Arrays 3