Wildcardmatching and regex, wildcard matching and regular expression matching

Wildcardmatching: Wildcard Match

Algorithm Analysis:

1. Two pointer I, J points to the string, the matching formula respectively.

2. If matched, advance directly with 2 pointers.

3. If the matching formula is *, match in the string.

Note the location where the previous comparison was recorded

Implement wildcard pattern matching with support for and ‘?‘‘*‘

‘?‘ Matches any single character.

' * ' Matches any sequence of characters (including the empty sequence).

Some examples:ismatch ("AA", "a") →falseismatch ("AA", "AA") →trueismatch ("AAA", "AA") →falseismatch ("AA", "*") → Trueismatch ("AA", "A *") →trueismatch ("AB", "? *") →trueismatch ("AaB", "C*a*b") →false

public class Wildcardmatching {public Boolean isMatch (string s, String p) {if (s = = NULL | | p = = NULL)        {return false;        } int indexs = 0;                int indexp = 0;        int lenS = S.length ();                int LENP = P.length ();         int PreP = -1;//Record wildcard as * when wildcard string subscript int preS = -1;//record wildcard to match string subscript while (Indexs < LenS) { Non-* matching, indexs and indexp simultaneously move if (Indexp < LENP && (P.charat (indexp) ==s.charat (indexs) | |             P.charat (indexp) = = '? '))                {indexp++;            indexs++;             }//Touch *, record at this time the subscript else if P,s (Indexp < LENP && P.charat (indexp) = = ' * ')                {PreS = Indexs;                                PreP = Indexp;            P Compare pointer back move, assuming * match empty string indexp++;          } else if (PreP! =-1)//try to match * {      Indexp = prep;//p fallback to * position indexp++;            pres++;//try to match *, move one character at a time, then compare P's * Trailing characters and s remaining characters, if not matched, fall back to the position of * Indexs = PreS;            } else {return false;                }} while (Indexp < LENP) {if (P.charat (indexp)! = ' * ') {            return false;        } indexp++;    } return true; }//through examples AaB and *ab to simulate the process public static void main (string[] args) {wildcardmatching wm = new wildcardmatching (); System.out.println (Wm.ismatch ("AaB", "*ab"));}}

Regex: Regular expression matching

Problem Description: 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). The function prototype should be:bool IsMatch (const char *s, const char *p) Some examples:ismatch ("AA", "a") →falseismatch ( "AA", "AA") →trueismatch ("AAA", "AA") →falseismatch ("AA", "A *") →trueismatch ("AA", ". *") →trueismatch ("AB", ". *") →true IsMatch ("AaB", "C*a*b") →true

Algorithm analysis:. * can match any string, such as AB match. *, not to say let. Match A and then go to match *, but * match is., i.e. (. *) = = (...), so. * Matches all strings.

Using dynamic programming, for match string p, we discuss three cases, p length is 0,p length is 1,p length is greater than 1 (P's second string is *,p second string is not *)

Dynamic planning public class Regex2 {public Boolean isMatch (string s, String p) {//P length is 0, boundary condition. if (p.length () = = 0) {return s.length () = = 0;}//P length is 1, boundary condition.        if (p.length () = = 1) {//S length is 0if (S.length () < 1) {return false;} There are two cases of first element matching//if P is. Then the first element and P must match, and if the first element of P is the same as the first element of S, it must match. else if ((P.charat (0)! = S.charat (0)) && (P.charat (0)! = '. ')) {return false;} Otherwise, in addition to the first matching element, compare other elements to the idea of dynamic planning. else {return IsMatch (s.substring (1), p.substring (1));}} The second element of P is not a *,* representing 0 or more preceding elements if (P.charat (1)! = ' * ') {if (S.length () < 1) {return false;} else if ((P.charat (0)! = S.charat (0)) && (P.charat (0)! = '. ')) {return false;} else {return IsMatch (s.substring (1), p.substring (1));}} Else//p's second element is that *{//* represents 0 preceding elements if (IsMatch (S, p.substring (2)) {return true;}//* represents one or more preceding elements//traversal s, if s's I element equals P's first element, or the first element of P is., the string that matches the third element of S i+1 and p for (int i = 0; I<s.length () && (S.charat (i) = = P.charat (0) | | (P.charat (0) = = '. ')); i + +) {if (IsMatch (s.substring (i + 1), p.substring (2))) {return true;}}return false;}} public static void Main (string[] args) {Regex2 reg2 = new Regex2 (); System.out.println (Reg2.ismatch ("Aaba", ". *"));}}

Wildcardmatching and regex, wildcard matching and regular expression matching