Task description

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.

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.

Assume that:

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

Complexity:

  • expected worst-case time complexity is O(log(N));
  • expected worst-case space complexity is O(1).
 
Solution

 
Programming language used: Java
Total time used: 8 minutes
Code: 09:57:26 UTC, java, final, score:  100
// 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"); class Solution {
public int solution(int N) {
// write your code in Java SE 8
int longest=0,temp=0;
int test = N;
boolean power = false;
if(test % 2 == 0){
test = test / 2;
power = true;
}
while(test !=0) {
if(test % 2 ==0) {
if(power) {
test = test / 2;
continue;
}
temp++;
} else {
temp = 0;
power =false;
}
test = test / 2;
if(longest < temp)
longest = temp;
}
return longest;
}
}
 
 

Codility---BinaryGap的更多相关文章

  1. [codility] Lession1 - Iterations - BinaryGap

    Task1: A binary gap within a positive integer N is any maximal sequence of consecutive zeros that is ...

  2. codility上的练习 (1)

    codility上面添加了教程.目前只有lesson 1,讲复杂度的……里面有几个题, 目前感觉题库的题简单. tasks: Frog-Jmp: 一只青蛙,要从X跳到Y或者大于等于Y的地方,每次跳的距 ...

  3. Codility NumberSolitaire Solution

    1.题目: A game for one player is played on a board consisting of N consecutive squares, numbered from ...

  4. codility flags solution

    How to solve this HARD issue 1. Problem: A non-empty zero-indexed array A consisting of N integers i ...

  5. GenomicRangeQuery /codility/ preFix sums

    首先上题目: A DNA sequence can be represented as a string consisting of the letters A, C, G and T, which ...

  6. *[codility]Peaks

    https://codility.com/demo/take-sample-test/peaks http://blog.csdn.net/caopengcs/article/details/1749 ...

  7. *[codility]Country network

    https://codility.com/programmers/challenges/fluorum2014 http://www.51nod.com/onlineJudge/questionCod ...

  8. *[codility]AscendingPaths

    https://codility.com/programmers/challenges/magnesium2014 图形上的DP,先按照路径长度排序,然后依次遍历,状态是使用到当前路径为止的情况:每个 ...

  9. *[codility]MaxDoubleSliceSum

    https://codility.com/demo/take-sample-test/max_double_slice_sum 两个最大子段和相拼接,从前和从后都扫一遍.注意其中一段可以为0.还有最后 ...

  10. *[codility]Fish

    https://codility.com/demo/take-sample-test/fish 一开始习惯性使用单调栈,后来发现一个普通栈就可以了. #include <stack> us ...

随机推荐

  1. [ACM] POJ 1094 Sorting It All Out (拓扑排序)

    Sorting It All Out Time Limit: 1000MS   Memory Limit: 10000K Total Submissions: 26801   Accepted: 92 ...

  2. 在Windows中安装MinGW-w64(有图,一步一步)

    在Windows中安装MinGW-w64 发表回复 如需配合Sublime Text 3编译C程序, 请参考本站文章: 使用Sublime Text 3与MinGW-w64编译C语言程序 MinGW, ...

  3. matlab 图像的保存

    gcf:获取当前显示图像的句柄: 默认 plot 的 position 是 [232 246 560 420] 0. save >> A = randn(3, 4); >> B ...

  4. Arcgis api for javascript学习笔记(4.5版本)-三维地图并叠加天地图标注

    1.三维地图实现 在官网的demo中就有三维地图的实现,如下图所示 <!DOCTYPE html> <html> <head> <meta charset=& ...

  5. GoldenGate过程 abend,报错OGG-00868 ORA-02396: Exceeded Maximum Idle Time, Please Connect Again

    GoldenGate过程 abend,报错OGG-00868 ORA-02396: Exceeded Maximum Idle Time, Please Connect Again 参考原始: Gol ...

  6. SQLServer2008-2012开启远程连接的配置方法

    一.远程连接端口设置(很关键的一步)1.在服务器上打开SQL Server Configuration Manager.选择SQL Server配置管理器->SQL Server 网络配置-&g ...

  7. 图像滤镜艺术---(Punch Filter)交叉冲印滤镜

    原文:图像滤镜艺术---(Punch Filter)交叉冲印滤镜 (Punch Filter)交叉冲印滤镜 本文介绍一种交叉冲印效果的代码实现,至于原理,不在累赘,直接看代码:  int f_TPun ...

  8. imp dll时遇见的非常恶心的问题

    我需要导入dll库中这样一个函数VM661JTCPDLL_API int admin_login(sel_admin_ret* sel_admins, int num, char* admin_nam ...

  9. BooleanToColorConverter

    public class BooleanToColorConverter : IValueConverter { public object Convert(object value, Type ta ...

  10. C#高性能大容量SOCKET并发(九):断点续传

    原文:C#高性能大容量SOCKET并发(九):断点续传 上传断点续传 断点续传主要是用在上传或下载文件,一般做法是开始上传的时候,服务器返回上次已经上传的大小,如果上传完成,则返回-1:下载开始的时候 ...