java - Comparing integers and creating the smallest integer possible from the digits of the given integers -

i need write following method: accepts 2 integer parameters , returns integer. if either integer not 4 digit number method should return smaller integer. otherwise, method should return 4 digit integer made of smallest digit in thousands place, hundreds place, tens place , ones place. cannot turn integers strings, or use lists, or arrays.

for example biggestloser(6712,1234) returns 1212 example biggestloser(19,8918) returns 19

here's how i've started write it:

public static int biggestloser(int a, int b){     if(a<9999 || b<9999){         if(a<b)             return a;         else if(b<a)             return b;     }     int at=a/1000;     int ah=a%1000/100;     int an=a%100/10;     int ae=a%10;     int bt=b/1000;     int bh=b%1000/100;     int bn=b%100/10;     int be=a%10;     if(at<bt && ah<bh && an<bn && ae<be)         return at*1000+ah*100+an*10+ae;     else if(at<bt && ah<bh && an<bn && be<ae)         return at*1000+ah*100+an*10+be;     else if(at<bt&& ah<bh && bn<an && ae<be)     else return at*1000+ah*100+bn*10+ae; 

however, looks i'm going have write way many if statements, there shorter way write code?

public static int biggestloser(int a, int b) {     if (a < 1000 || >= 10000 || b < 1000 || b >= 10000) {         return math.min(a, b);     } else {         // both , b 4 digits         int result = 0 ;         int multiplier = 1 ;          (int digit = 0; digit < 4; digit++) {             int nextdigit = math.min(a % 10, b % 10);             result = result + nextdigit * multiplier  ;             multiplier = multiplier * 10 ;             = / 10 ;             b = b / 10 ;         }          return result ;     } } 

how work? a % 10 remainder when a divided 10: in other words least significant digit of a (the "ones place").

a = / 10 performs integer division, divides a 10 , ignores fraction. 1234 becomes 123, , on next iteration 123 becomes 12, etc. in other words, discards "ones place".

so first time through loop, @ "ones" a , b, find smallest one, , add result. drop "ones" both a , b. used "tens" "ones". second time through loop, smallest "ones" again: smallest "tens". want add result, need multiply 10. multiplier: each time through loop multiplier multiplied 10. each time, smallest "ones", multiply correct thing, add result, , drop "ones" a , b.

just fun, here's implementation needs 1 statement (and works if replace "four digits" positive number of digits). can ask instructor explain ;).

public static final int num_digits = 4 ; public static final int max = (int) math.pow(10, num_digits) ; public static final int min = max / 10 ;   public static int biggestloser(int a, int b) {     return (a < min || >= max || b < min || b >= max) ? math.min(a, b) :         intstream.iterate(1, multiplier -> multiplier * 10).limit(num_digits)             .map(multiplier -> math.min((a / multiplier) % 10, (b / multiplier) % 10) * multiplier )             .sum(); }