문제

https://app.codility.com/programmers/lessons/1-iterations/binary_gap/

A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both ends in the binary representation of N.

For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no binary gaps. The number 32 has binary representation 100000 and has no binary gaps.

Write a function:

class Solution { public int solution(int N); }

that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N doesn't contain a binary gap.

For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation '100000' and thus no binary gaps.

Write an efficient algorithm for the following assumptions:

N is an integer within the range [1..2,147,483,647].

풀이

• 이진법으로 나타냈을 때 1과 1사이의 0이 연속으로 존재할 때의 0의 개수 중 최대값을 구하라

코드

// you can also use imports, for example:
// import java.util.*;

// you can write to stdout for debugging purposes, e.g.
// System.out.println("this is a debug message");

/**
이진법으로 바꿨을 때 양 옆에 1을 두고 연속된 0의 개수의 최대값
만약 없는 경우 0 리턴
**/
class Solution {
    public int solution(int N) {
        // write your code in Java SE 8

        String binaryStr = "";
        int tmp = N;

        while(tmp > 0){
            binaryStr += Integer.toString(tmp%2);
            tmp /= 2;
        }

        int answer = 0;
        int cnt = 0;
        
        boolean check = false; // 1이 시작 됐는지 체크
        for(char c : binaryStr.toCharArray()){
            if(check == false && c == '1'){
                // 1이 시작되는 경우
                check = true;                
            }
            else if(check == true && c == '1'){
                // 양 옆이 1인 경우의 마지막 1이 나온 경우
                answer = Math.max(answer, cnt);
                cnt = 0; // 연속된 0이 끝났으므로 개수 초기화
            }
            else if(check == true && c == '0'){
                cnt++; // 처음에 1이 시작됐고 0이 나온 경우
            }
        }

        return answer;
    }
}