Count maximum consecutives 1's in binary number

Question

How can I count in Java the maximum consecutive 1's from a binary number? For instance, the user enters an integer like 13 this would be in binary 1101. The binary number 1101 has 2 consecutive one's, so the output should be 2. Another example the number 5 would be in binary 0101, so the output should be 1. My program doesn't work correctly if I type the number 439.

Scanner scanner = new Scanner(System.in);
        int numb = scanner.nextInt();
        int tempCount = 0;
        int count = 0; 

        // numb = 439 -> output 3
        // numb = 13  -> output 2
        // numb = 1   -> output 1 
        while (numb >= 1) {

            if (numb % 2 != 0) {

                tempCount++;
                numb /= 2;
                count = tempCount;
                if (numb % 2 != 0) {
                    tempCount++;
                    count = tempCount;
                    numb /= 2;
                }

            }
            else {
                numb /= 2;
                tempCount = 0;
            }
        }
        System.out.println(count);

Answer 1

I figured it out, here is the solution. I have forgotten an nested if statement.

Scanner scanner = new Scanner(System.in);
int numb = scanner.nextInt();
int tempCount = 0; //save temporarily the longest sequence of one's in this variable 
int count = 0; // before the sequence of one's gets interrupted, I save the value from tempCount in count

// numb = 439 -> output 3
// numb = 13  -> output 2
// numb = 1   -> output 1 
while (numb >= 1) {

    if (numb % 2 != 0) {

        tempCount++;
        numb /= 2;
        if(count < tempCount){
            count = tempCount;
        }

    }
    else {
        numb /= 2;
        tempCount = 0;
    }
}
System.out.println(count);