https://app.codility.com/demo/results/trainingV47U7M-RBU/

문제

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].
Copyright 2009–2023 by Codility Limited. All Rights Reserved. Unauthorized copying, publication or disclosure prohibited.

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");
import java.util.*;
class Solution {
    public int solution(int N) {
    Stack<Integer> st = new Stack<Integer>();
    ArrayList<Integer> bnlist = new ArrayList<Integer>();

    while(N>=1){
        st.push(N%2);
        N=N/2;
    } 

    int bgLen = 0;
    int current = 0;

    while(st.size()!=0){
        current = st.peek();
        st.pop();
        if(current ==0){
            bgLen++;
        }
        else{
            bnlist.add(bgLen); 
            bgLen = 0;
        }
    }
    Collections.sort(bnlist, Collections.reverseOrder());
    
    return bnlist.get(0);
    }
}

풀이

Stack에 이진수를 만들어서 담고,
Stack에서 peek 한 값을 current에 저장한 후 pop한다.
current가 0이면 bgLen값을 늘리고,
1이면 bgLen값을 리스트에 저장한 후 다시 0으로 초기화 해준다.
리스트에는 1과 1사이에 있는 0의 갯수가 담기게 되고, 리스트의 가장 큰 수를 return 한다.