java - Extracting even digits from int -

here problem solving.

write method evendigits accepts integer parameter n , returns integer formed removing odd digits n. following table shows several calls , expected return values:

call    valued returned evendigits(8342116);    8426 evendigits(4109);   40 evendigits(8);  8 evendigits(-34512); -42 evendigits(-163505);    -60 evendigits(3052);   2 evendigits(7010496);    46 evendigits(35179);  0 evendigits(5307);   0 evendigits(7);  0 

if negative number digits other 0 passed method, result should negative, shown above when -34512 passed.

leading zeros in result should ignored , if there no digits other 0 in number, method should return 0, shown in last 3 outputs.

i have far -

public static int evendigits(int n) {     if (n != 0) {          int new_x = 0;         int temp = 0;         string subs = "";     string x_str = integer.tostring(n);         if (x_str.substring(0, 1).equals("-")) {              temp = integer.parseint(x_str.substring(0, 2));              subs = x_str.substring(2);         } else {              temp = integer.parseint(x_str.substring(0, 1));              subs = x_str.substring(1);         }          if (subs.length() != 0) {              new_x = integer.parseint(x_str.substring(1));         }                  if (temp % 2 == 0) {              return integer.parseint((integer.tostring(temp) + evendigits(new_x)));         } else {             return evendigits(new_x);         }     }     return 0; } 

why people seem want convert string deal digits? java has arithmetic primitives handling job. example:

public static int evendigits(int n) {     int rev = 0;     int digitcount = 0;      // handle negative arguments     if (n < 0) return -evendigits(-n);      // extract digits variable rev     while (n != 0) {         if (n % 2 == 0) {             rev = rev * 10 + n % 10;             digitcount += 1;         }         n /= 10;     }      // digits extracted in reverse order; reverse them again     while (digitcount > 0) {         n = n * 10 + rev % 10;         rev /= 10;         digitcount -= 1;     }      // result in n     return n; } 

although makes no difference simple academic exercise such one, handling job via arithmetic alone can expected perform better involving converting string.