write a function to check given string matches with given pattern

write a function to check given string matches with given pattern
Condition: only one wildcard used in the pattern, that is '*', but can be used in the pattern more than once.
Example: 

pattern: *abc*def*.doc*
string: adsfabcxyzdefgh.docx
output : Matching


pattern: *abc*def*.doc*
string: adsfabcxyzdefgh.do1cx
output : Not Matching

public class isMatchingWithWildcard {

public static void main(String [] args){

String pattern = "*abc*def*.doc*";

String str = "adsfabcxyzdefgh.do1docx";

if(isMatching(str, pattern))

System.out.print("Matching");

else

System.out.print("Not Matching");

}

public static Boolean isMatching(String str, String pattern){

int l = pattern.length();

if(pattern.lastIndexOf("*") == l-1)

pattern= (String) pattern.subSequence(0, l-1);

if(pattern.charAt(0) == '*')

pattern= (String) pattern.subSequence(1, pattern.length());

pattern = pattern.replace("*", "__");

String [] patternArray = pattern.split("__");

String sorttenString =str;

for(String aPattern:patternArray){

String [] temp = sorttenString.split(aPattern);

if(temp.length==1 && (aPattern == patternArray[patternArray.length-1] 

&& !sorttenString.toLowerCase().contains(aPattern.toLowerCase())))

return false;

str.substring(temp[0].length()+aPattern.length());

}

return true;

}

}

Let me know your thoughts

Find the seed of a number.

Eg : 1716 = 143*1*4*3 =1716so 143 is the seed of 1716.
find all possible seed for a given number.

// Input   ||  OUTPUT 
//  1250        125       
//  174720      1456   

#include<stdio.h>
int main()
{
    int n,i,j,m,f=0;
    scanf("%d",&n);
    printf("\n");
    for(i=0;i<n;i++)
    {
        m=i;
        j=i;
        while(j)
        {
            m=m*(j%10);
            j=j/10;
        }
        if(m==n)
        {
            printf("\n%d",i);
            f=1;
        }
    }
    if(!f)
        printf("\nNo such number exit\n\n");
return 0;
}