CareerCup

public int solution(int N) {
		// write your code in Java SE 8
		String binary = Integer.toBinaryString(N);
		int compte = 0, comptegap = 0, result = 0;
		char[] value = binary.toCharArray();

		System.out.println(value);

		for (char v : value) {
			if (v == '1') {
				compte++;
			}

			if (v == '0') {

				if (compte == 2) {

					if (result < comptegap) {
						result = comptegap;
					}

					comptegap = 0;
				} else {
					result = 0;
				}

				comptegap++;
			}
		}

		return result;
	}

We do following

public static int BinaryGap(int n)
{
	int maxGap = 0;
	int currentGap = 0;
	while(n > 0)
	{
		if(n % 2 == 0)
		{
			currentGap++;
			if(maxGap < currentGap)
			{
				maxGap = currentGap;
			}
		}
		else if(n % 2 == 1)
		{
			currentGap = 0;
		}
		n = n/2;
	}
	return maxGap;
}

In a loop, divide the number by 2. Keep count of the number of times, n%2==0 is found continuously

public int maxGap(int n){
	if(n<0){ return -1;}
	int maxgap = 0;
	int gap=0;
	while(n>0){
		if(n%2==0){
			gap++;
			if(gap>maxgap){ maxgap = gap; }
		} else {
			gap=0;
		}
	}
}

ujyujy

public int solution(int N) {
String binary = Integer.toBinaryString(N);
int compte = 0, comptegap = 0, result = 0;
char[] value = binary.toCharArray();

System.out.println(value);

for (char v : value) {
if (v == '1') {
compte++;
}

if (v == '0') {

if (compte == 2) {

if (result < comptegap) {
result = comptegap;
}

comptegap = 0;
} else {
result = 0;
}

comptegap++;
}
}

return result;
}