Add document for :https://www.isharkfly.com/t/binary-gap/311
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- [下一个斐波拉契数](/algorithm/next-fibonacci-number.md)
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- [下一个斐波拉契数](/algorithm/next-fibonacci-number.md)
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- [一个字符串包裹函数](/algorithm/a-word-wrap-functionality.md)
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- [一个字符串包裹函数](/algorithm/a-word-wrap-functionality.md)
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- [打印 100 以内的素数](/algorithm/prime-numbers-from-1-to-100.md)
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- [打印 100 以内的素数](/algorithm/prime-numbers-from-1-to-100.md)
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- [二进制空白](/algorithm/binary-gap.md)
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# Binary Gap(二进制空白)
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> 🔔 参与讨论:https://www.isharkfly.com/t/binary-gap/311
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**英文描述**
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A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is surrounded by ones at both
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ends in the binary representation of N.
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For example, number 9 has binary representation 1001 and contains a binary gap of length 2. The number 529 has binary
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representation 1000010001 and contains two binary gaps: one of length 4 and one of length 3. The number 20 has binary
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representation 10100 and contains one binary gap of length 1. The number 15 has binary representation 1111 and has no
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binary gaps. The number 32 has binary representation 100000 and has no binary gaps.
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Write a function:
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class Solution { public int solution(int N); }
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that, given a positive integer N, returns the length of its longest binary gap. The function should return 0 if N
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doesn't contain a binary gap.
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For example, given N = 1041 the function should return 5, because N has binary representation 10000010001 and so its
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longest binary gap is of length 5. Given N = 32 the function should return 0, because N has binary representation '
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100000' and thus no binary gaps.
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Write an efficient algorithm for the following assumptions:
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N is an integer within the range [1..2,147,483,647].
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**中文描述**
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这里我不按照原文一字一字的翻译,但是尽量按照题目的要求把题目解释清楚。
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这里题目的要求是,将 N 为一个整数类型的数据,转换为一个 2 进制的字符串,然后在返回的字符串中返回最大的 0 的间隔数字。
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例如 529 转换为 2 进制的字符串为:1000010001,在这里,将会存在以 1 为分割的字符串 0000 和 000,这 2 个字符串的长度分别为 4 和
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3。
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我们的算法需要返回的值为 4。
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### 思路和点评
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这个题目的思路其实比较简单,你需要首先将 N 这个整数,转换为 0 和 1 的字符串。然后在转换成功的字符串中返回以 1 分分割的 0
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的长度。
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这里可能需要考虑下面的几种情况。
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| 情况 | 结果 |
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|----|---------------------------------|
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| 11 | 这个情况应该返回的长度为 0 |
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| 10 | 这个情况因为没有被 1 这个字符串封闭,因此应该返回长度为 0 |
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传统的思路应该是采取字符串分割的方式,进行遍历后获得结果。
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我们在这里采取一种相对不是非常常规的方式,例如在 10000010001 字符串中插入 #,将字符串变为 #1#00000#1#000#1#。
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然后将字符串按照 1 进行分割,那么分割后的数组应该分别存储的数据为:#,#0000#,#000#,#
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这里我们只需要找到 #...# 中值最大的连续 0 字符串就可以了。基本上可以使用 1 个字符串替换函数和一个字符串分割函数就可以了,并不需要多次存储和遍历。
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**源代码**
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源代码和有关代码的更新请访问 GitHub:
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https://github.com/cwiki-us/java-tutorial/blob/master/src/test/java/com/ossez/lang/tutorial/tests/codility/CodilityBinaryGapTest.java
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代码思路请参考:
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```java
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package com.ossez.lang.tutorial.tests.codility;
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import org.junit.Test;
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import org.slf4j.Logger;
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import org.slf4j.LoggerFactory;
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/**
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* <p>
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* More details about question see link below
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* <ul>
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* <li>@see <a href= "https://www.cwiki.us/display/ITCLASSIFICATION/Binary+Gap">https://www.cwiki.us/display/ITCLASSIFICATION/Binary+Gap</a>
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* </li>
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* </ul>
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* </p>
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*
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* @author YuCheng
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*
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*/
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public class CodilityBinaryGapTest {
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private final static Logger logger = LoggerFactory.getLogger(CodilityBinaryGapTest.class);
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/**
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*
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*/
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@Test
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public void testMain() {
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logger.debug("BEGIN");
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int N = 529;
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String intStr = Integer.toBinaryString(N);
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intStr = intStr.replace("1", "#1#");
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String[] strArray = intStr.split("1");
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int maxCount = 0;
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for (int i = 0; i < strArray.length; i++) {
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String checkStr = strArray[i];
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int countLength = 0;
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if (checkStr.length() > 2 && checkStr.startsWith("#") && checkStr.endsWith("#")) {
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checkStr = checkStr.replace("#", "");
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countLength = checkStr.length();
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if (maxCount < countLength) {
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maxCount = countLength;
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}
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}
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}
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logger.debug("MAX COUNT: [{}]", maxCount);
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}
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}
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```
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