Java program to find smallest number?

Question

Java

A simple task that led me a little into a stupor.

public static void main(String[] args) {
    int a, b, c, d;
    int min = 0;
    a = 12;
    b = 22;
    c = 33;
    d = 11;
    
    if(a>b){
      min = b;
    }
    if(min>c){
      min = c;
    }
    if(min>d){
      min = d;
    }
    System.out.println("Наименьшее число "+min);
  }

The problem is that if the first if a>b does not work, then min will not be filled with a number and by default it will be 0 , therefore all subsequent manipulations are useless, min will always be 0.
If min is not assigned a default value, then the code will not work at all.
How to find the smallest number?

Answer 1

int min = a;

Answer 2

1) Put the values ​​in an array
2) Walk through the array, remembering the minimums

min = array[0];
for (int i=1; i < array.length; i++) {
  if (array[i+1] < min)
    min = array[i+1]    
}

PS Correct me if anything, I do not write in Java.

Answer 3

int[] numbers = {9,8,7,6,5,4,3,2,1} ;
int min = numbers[0];
for ( int n : numbers ) {
     min = min < n ? min : n;
}
System.out.println("Min = " + min);

And a cheating brake method in two lines:

int[] numbers = {9,8,7,6,5,4,3,2,1} ;
Arrays.sort(numbers);
int min = numbers[0];
System.out.println("Min = " + min);