You don't need to reverse the actual binary string. You can just calculate the index.

String str = Integer.toBinaryString(num);
int len = str.length();
List<Integer> list = new ArrayList<>();
for (int i=0; i < len; i ++) {
  if (str.charAt(i) == '1') list.add(len - 1 - i);
}

You can just test the bits without turning the integer into a string:

List<Integer> onePositions(int input) {
  List<Integer> onePositions = new ArrayList<>();
  for (int bit = 0; bit < 32; bit++) {
    if (input & (1 << bit) != 0) {
      onePositions.add(bit + 1); // One-based, for better or worse.
    }
  }
  return onePositions;
}

Bits are usually counted from right to left, the rightmost bit being bit 0. The operation 1 << bit gives you an int whose bit numbered bit is set to 1 (and the rest to 0). Then use & (binary and) to check if this bit is set in the input, and if so, record the position in the output array.

Just check the bits in turn:

List<Integer> bits(int num) {
  List<Integer> setBits = new ArrayList<>();
  for (int i = 1; num != 0; ++i, num >>>= 1) {
    if ((num & 1) != 0) setBits.add(i);
  }
  return setBits;
}

Online Demo

6 [2, 3]
7 [1, 2, 3]
8 [4]

Just use the indexOf function of the String class

public class TestPosition {
    public static void main(String[] args) {
        String word = "110"; // your string
        String guess = "1"; // since we are looking for 1
        int totalLength = word.length();
        int index = word.indexOf(guess);
        while (index >= 0) {
            System.out.println(totalLength - index);
            index = word.indexOf(guess, index + 1);
        }
    }
}

You can use this solution:

    static List<Integer> convert(int input) {
        List<Integer> list = new ArrayList<>();
        int counter = 1;
        int num = (input >= 0) ? input : Integer.MAX_VALUE + input + 1;
        while (num > 0) {
            if (num % 2 != 0) {
                list.add(counter);
            }
            ++counter;
            num /= 2;
        }
        return list;
    }

It outputs:

[2, 3]
[1, 2, 3]
[4]

I would definitely prefer Andy's answer myself, even if it seems cryptic at first. But since no one here has an answer with streams yet (even if they are totally out of place here):

public List<Integer>  getList(int x) {
    String str = Integer.toBinaryString(x);
    final String reversed = new StringBuilder(str).reverse().toString();
    return IntStream.range(1, str.length()+1)
            .filter(i -> reversed.charAt(i-1)=='1')
            .boxed()
            .collect(Collectors.toList());
}

Since you wrote "modern Java", this is how it can be done with streams (Java 8 or better):

final int num = 7;

List<Integer> digits = IntStream.range(0,31).filter(i-> ((num & 1<<i) != 0))
        .map(i -> i+1).boxed().collect(Collectors.toList());

The map is only needed since you start counting at 1 and not at 0.

Then

System.out.println(digits);

prints

[1, 2, 3]

A silly answer, just for variety:

BitSet bs = BitSet.valueOf(new long[] {0xFFFFFFFFL & input});
List<Integer> setBits = new ArrayList<>();
for (int next = -1; (next = bs.nextSetBit(next + 1)) != -1;) {
  setBits.add(next + 1);
}

(Thanks to pero_hero for pointing out the masking was necessary on WJS's answer)

just for fun:

Pattern one = Pattern.compile("1");
List<Integer> collect = one.matcher(
             new StringBuilder(Integer.toBinaryString(value)).reverse())
            .results()
            .map(m -> m.start() + 1)
            .collect(Collectors.toList());
System.out.println(collect);

Given the original integer returns a list with the bit positions.

static List<Integer> bitPositions(int v) {
     return BitSet.valueOf(new long[]{v&0xFF_FF_FF_FFL})
                .stream()
                .mapToObj(b->b+1)
                .collect(Collectors.toList());
}

Or if you want to do bit shifting.

static List<Integer> bitPositions(int v ) {
    List<Integer> bits  = new ArrayList<>();
    int pos = 1;
    while (v != 0) {
        if ((v & 1) == 1) {
            bits.add(pos);
        }
        pos++;
        v >>>= 1;
    }
    return bits;

}

or if you want:

String strValue = Integer.toBinaryString(value);
List<Integer> collect2 = strValue.codePoints()
           .collect(ArrayList<Integer>::new,
                   (l, v) -> l.add(v == '1' ? strValue.length() - l.size() : -1), 
                   (l1, l2) -> l1.addAll(l2)).stream()
           .filter(e -> e >= 0)
           .sorted()
           .collect(toList());

May I propose a pure bit-wise solution?

static List<Integer> onesPositions(int input)
{
    List<Integer> result = new ArrayList<Integer>(Integer.bitCount(input));

    while (input != 0)
    {
        int one = Integer.lowestOneBit(input);
        input = input - one;
        result.add(Integer.numberOfTrailingZeros(one));
    }

    return result;
}

This solution is algorithmically optimal:

1. Single memory allocation, using Integer.bitCount to appropriately size the ArrayList in advance.
2. Minimum number of loop iterations, one per set bit¹.

The inner loop is rather simple:

• Integer.lowestOneBit returns an int with only the lowest bit of the input set.
• input - one "unset" this bit from the input, for next iteration.
• Integer.numberOfTrailingZeros count the number of trailing zeros, in binary, effectively giving us the index of the lowest 1 bit.

¹ It is notable that this may not be the most optimal way once compiled, and that instead an explicit 0..n loop based on the bitCount would be easier to unroll for the JIT.

a stream version of @Matthieu M. answer:

 List<Integer> list = IntStream.iterate(value, (v) -> v != 0, (v) -> v & (v - 1))
                .mapToObj(val -> Integer.numberOfTrailingZeros(val) + 1)
                .collect(toList());