02. Codility Lession 1. Binary Gap

문제내용

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1) 원문

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].

 

2) 간단 해석

주어진 10진수를 2진수로 바꿨을 때 1과 1 사이에 존재하는 0의 최대 갯수 구해야 한다.

(예: 529를 2진수로 바꾸면 1000010001 이고 1과 1 사이에 존재하는 0의 갯수는 4와 3이다.

     이때 최대 갯수인 4를 return해야 하는 메소드를 짜야 한다.)

 

 

풀이내용

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1) 1차 풀이

import java.util.*;

class Solution {
    public int solution(int N) {
        int gap = 0;
        
        // 1. make binary string
        String binary = makeBinary(N);

        // 2. replace 0 with the number of 0 and get the maximal number of 0
        gap = getMaxGap(binary);
        
        return gap;
    }

    public String makeBinary(int N) {
        List<Integer> list = new ArrayList<Integer>();
        String binary = "";

        while (N != 1) {
            list.add(N % 2);
            N = N / 2;
        }

        if (N == 1) {
            list.add(N);
        }
        Collections.reverse(list);
        
        for (int i : list) {
            binary += String.valueOf(i);
        }
        return binary;
    }

    public int getMaxGap(String binary) {
        String tmp = "";
        int gap = 0, cnt = 0;

        if (binary.contains("0")) {
            for (char c : binary.toCharArray()) {
                if (c == '1') {
                    if (cnt == 0) {
                        tmp += String.valueOf(c);    // add 1
                    } else {
                        tmp += String.valueOf(cnt);   // add the number of 0
                        tmp += String.valueOf(c);     // add 1
                        cnt = 0;    // re-initialize cnt
                    }
                } else {    // if c is 0
                    cnt++;
                }
            }
            tmp = tmp.replace("1", "");

            if (!tmp.equals("")) {
                List<Integer> list = new ArrayList<Integer>();
                
                for (char c : tmp.toCharArray()) {
                    System.out.println("c: " + c);
                    list.add(Character.getNumericValue(c));
                }
                
                if (list.size() != 0) {
                    gap = Collections.max(list);
                }
            }
        }

        return gap;
    }
}

 

2) 1차 풀이 결과

33점

아니 이보시오 코딜리티 양반!! 내가 33점이라니!!!!

 

분석 총평은 아래와 같다 :< 

오늘은 늦었으니 내일 다시 수정해봐야지!!