Leetcode 44.Wildcard Matching (wildcard match) ideas and methods for solving problems

Wildcard Matching

‘?‘ Matches any single character.
' * ' Matches any sequence of characters (including the empty sequence).

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") →false
IsMatch ("AA", "AA") →true
IsMatch ("AAA", "AA") →false
IsMatch ("AA", "*") →true
IsMatch ("AA", "A *") →true
IsMatch ("AB", "? *") →true

IsMatch ("AaB", "C*a*b") →false

Train of thought: this difficulty is bigger, this difficulty refers to the difficulty of time complexity, I use recursive write code, correctness is no problem, but time efficiency simply can not endure. The submission was decisive, however.

My code is as follows (for reference only):

public class Solution {public    Boolean isMatch (string s, String p) {    //Boundary condition to determine        if (s.length () = = 0)            return P.length () = = 0;        if (p.length () = = 0)            return false;        If * If            (p.charat (0) = = ' * ') {            int i = 1;            while (I < p.length () && p.charat (i) = = ' * ') {                i++;            }            p = p.substring (i);//Remove the * if it also matches            i = 0;            while (s.length () > 0) {                if (IsMatch (s,p))//because * matches any sequence, p matches any s to                    return true;                s = s.substring (1,s.length ());//string 1 per intercept, try each            }            return IsMatch (s,p);//If none match, then S is empty        }else{//if not * See if the first character of S matches if            (P.charat (0) = = '? ' | | S.charat (0) = = P.charat (0))                return IsMatch (s.substring (1,s.length ()), P.substring (1,p.length ()));            Return false;//mismatch returns false        }}    }

In the forum to see the great God code, unfortunately when the idea did not understand, left to learn it later.

The code is as follows:

public class Solution {public Boolean IsMatch (string s, String p) {/** * This is to see the forum people with the idea of DP realized * I did not realize, with the recursive write although correct,        But tle * This question I also did not thoroughly understand this code, can only write here, later come to study * * int m = s.length (), n = p.length (); int count = 0;//* number for (int i = 0; i < n; i++) {if (P.charat (i) = = ' * ') count++;//statistics * The number of} if (Count==0 && m! = N) return false;//if not * and m!=n, certainly does not match the else if (N-count > m) return false;//if not the number of * >m, then equal to no * or greater than M, mismatch boolean[] match = new boolean[m+1];//default is all false match                [0] = true;//The first set to True for (int i = 0; i < n; i++) {if (P.charat (i) = = ' * ') {//= *, then all data after s corresponds to default match                 for (int j = 0; J < m; J + +) {match[j+1] = Match[j] | | match[j+1]; }} else {//if it is not *, use the last match for (int j = m-1; J >= 0; j--) {match[j+1] = ( P.charat (i) = = '? ' | | P.charat (i) = = S.charat (j)) && Match[j];            } Match[0] = false;    }} return match[m]; }}

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Leetcode 44.Wildcard Matching (wildcard match) ideas and methods for solving problems