Since we recently start using docker for product release, I’m dealing with a lot of problems related with docker. One of the headache is sometimes, the running application performs different in docker containers with the one directly runs in local without docker. For me, a more specific example is my webApp, running in gradle with Jetty9 plugins, can finish our load test perfectly only on my ubuntu machine. But it will get memory leak error when runs in docker container. So I decided to look into it by Java profiler.

It’s quite easy to use a GUI profiler to monitor the whole process how an application uses CPU and memory during running time. But it’s a little complicated to attach it in a container. Firstly I tried to use YourKit by its remote profiling option. The basic steps are:

1. Modify Dockerfile to get YourKit installed.

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   RUN apt-get update RUN apt-get install -y wget RUN wget https://www.yourkit.com/download/yjp-2016.02-b29-linux.tar.bz2 RUN tar xfj yjp-2016.02-b29-linux.tar.bz2 "-agentpath:/usr/share/jetty/yjp-2016.02/bin/linux-x86-64/libyjpagent.so=disablestacktelemetry,disableexceptiontelemetry,delay=10000,sessionname=Jetty", \ EXPOSE 10001

2. After building the image, access into the docker container and attach to JVM:

   1 2	docker exec -i -t <containerID> /bin/bash <directory with unpacked content>/bin/yjp.sh -attach

3. In YourKit GUI, link to the host:port.
   But every time when I try to attach to JVM (step 2), I get error below:

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   com.sun.tools.attach.AttachNotSupportedException: Unable to open socket file: target process not responding or HotSpot VM not loaded.

   It should be solved by just switching to the user created the container. But it didn’t work. So I tried to use JProfiler.
   Here are the steps:
4. Modify the Dockerfile to get JProfiler installed:

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   RUN \ apt-get update && \ apt-get install -y wget && \ wget http://download-keycdn.ej-technologies.com/jprofiler/jprofiler_linux_9_1_1.tar.gz && \ tar -xzf jprofiler_linux_9_1_1.tar.gz -C /usr/local ENV JPAGENT_PATH="-agentpath:/usr/local/jprofiler9/bin/linux-x64/libjprofilerti.so=nowait" EXPOSE 8849

   Be careful of the version. The one installed in docker container should be the same version as the one you use to monitor.
5. Execute the command in docker container:

   1 2	docker exec -i -t <containerID> /bin/bash /usr/local/jprofiler9/bin/jpenable

   That’s it. Now we can link to the remote from anywhere by host:port. The default port is 8849.

But the sad story is after looking into the result of the profiler, I still don’t know how it got memory leaked. Here is a discussion about the similar docker problem, maybe I’m not the only one.
I’ll update once I figured out…

Spend a whole night working on updating Hexo. It is much more complicated than I expected. For anyone’s convenience, I record some key points when migrate from Hexo 2.x to 3.0 besides the official documentation.

First things first: uninstall Hexo

Yes. Though the official wiki says that we can directly migrate from 2.x to 3.0, this doesn’t work (to me). Besides, instead of wasting time figuring out what’s wrong when updating, re-installating and configurating Hexo 3.0 seems acceptable to me. You can try to update directly but I didn’t do that successfully.
Before your uninstallation, backup your posts and any other custom setting files (eg. _config.yml, your own theme file, etc).
To uninstall Hexo, use the node package manager(npm):

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npm uninstall hexo 

Now, install Hexo 3.0

Run

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$ sudo npm install -g hexo-cli

Hexo wiki doesn’t mention the administration permission, but in my experience we have to add sudo here.

Then we need to setup hexo folder in your desired place(folder).

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$ hexo init <folder>
$ cd <folder>
$ sudo npm install

Still, try sudo when you fail to run any command.

For your future convenience, we install all necessary services here:

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npm install hexo-generator-index --save
npm install hexo-generator-archive --save
npm install hexo-generator-category --save
npm install hexo-generator-tag --save
npm install hexo-server --save
npm install hexo-deployer-git --save
npm install hexo-deployer-heroku --save
npm install hexo-deployer-rsync --save
npm install hexo-deployer-openshift --save

and any plugins you probably need:

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npm install hexo-renderer-marked@0.2 --save
npm install hexo-renderer-stylus@0.2 --save
npm install hexo-generator-feed@1 --save
npm install hexo-generator-sitemap@1 --save

All in sudo mode (or you can directly do all of these in sudo -i mode).
Now, we finished installation.

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Introduction

ASCII

Single byte encoding only using the bottom 7 bits(0-127). The top bit is always 0.

In English, 128 symbols are enough to represent all character. But in other situations, French for example, they are insufficient. So we use the top bit to represent accents so that there are up to 256 characters. ‘é’ encodes as 1000 0010(130).
But here comes another problem. In different languages, the same binary encoding represents different characters, such as 130 in French is é, but in Hebrew, it is Gimel(ג). Not to mention Chinese characters(more than 100 thousand). So we introduce another encoding system, unicode.

Unicode

“Unicode encoding” is more properly known as UTF-16: 2 bytes per “code point”. This is the native format of strings in .NET. Values outside the Basic Multilingual Plane(BMP) are encoded as surrogate pairs. These are relatively rarely used - which is a good job, as very few developers get them right, I suspect. “Unicode” is really the character set - it is unfortunate that the term is also used as a synonym for UTF-16 in .NET and various Windows applications.
Unicode can be implemented by different character encodings. The most commonly used encodings are UTF-8, UTF-7 and UTF-32:
UTF-8: Variable length encoding, 1-4 bytes covers every current character. ASCII values are encoded as ASCII.
UTF-7: Usually used for mail encoding. Chances are if you think you need it and you’re not doing mail, you’re wrong. (not widely used at all.)
UTF-32: Fixed width encoding using 4 bytes per code point. This isn’t very efficient, but makes life easier outside the BMP.

UTF-8

UTF-8 has become the dominant character encoding for the World Wide Web. The rule of UTF-8 is:
(1) If this is in ASCII, UTF-8 is the same with ASCII
(2) For n UTF bytes(n > 1), the first n bits in the first byte set as 1, the n + 1 bit sets as 0, all the first two bits in the following bytes are all 10. The rest of the bits are represented as the unicode of the character.

For example, the unicode of “严” is 4E25 (100111000100101). According to the table above, this character belongs to the third row. So its UTF-8 is E4B8A5 (11100100 10111000 10100101).
We can see that actually unicode is different with UTF-8. But there are some libraries that can do the convert.

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Single Number

XOR features:
a ^ b = c
a ^ c = b
b ^ c = a
a ^ 0 = a
a ^ a = 0
(a ^ b) ^ c = a ^ (b ^ c)

• I: all the numbers appear twice except one
  XOR all the numbers, and the result is the one
• II: all the numbers appear three times except one
  Use a int[32] to store each bit, if the number of this bit can be mod by 3, we set it as 0, otherwise we set it as 1.
• III: all the numbers appear twice except two
  XOR all the numbers so we can get the two by XOR. Since these two elements are not the same, there is at least one bit that different. We find this different position and XOR all the elements which have 1 on position different, the result is one of the two. Finally, we XOR the found one with XOR-ed two, we can get another one.

Majority Number

• I: find the number appears more than half
  Since it is a strict majority number, we only need to maintain a variable result and a counter. Initially, when the counter faces with a different number, minus one of counter and change the result to this new number, otherwise we only need to add 1 to the counter. And finally, the result we maintained is the result. (We need to do another pass to make sure this number is the one we are looking for)
• II: find the number appears more than 1/3
  Similar with I, we maintain two variables and two counters so that we can get most appeared two numbers. And then we put these two numbers back to the array to find out which one appears more than 1/3. (We can’t use counter to find out, because counters have minus operation during the traversal)
• III: find the number appears more than 1/k
  We maintain a HashMap<number, counter>. Since if majority, each k different numbers should have more than one element. When we get k entries in the map, we remove entries which values are 1. Finally, find the entry that have highest value(count), that key should be the majority number. In this way, we can implement this in O(n) time and O(k) extra space.

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  public class Solution { public int majorityNumber(ArrayList<Integer> nums, int k) { if(nums == null || nums.size() == 0) { return -1; } HashMap<Integer, Integer> map = new HashMap<Integer, Integer>(); for(int num : nums) { if(map.containsKey(num)) { map.put(num, map.get(num) + 1); } else { // if there are k entries, check if there are any entries that // its value == 1, that is not satisfied with "more than 1/k" if(map.size() == k) { Iterator<Map.Entry<Integer, Integer>> iter = map.entrySet().iterator(); while(iter.hasNext()) { Map.Entry<Integer, Integer> entry = iter.next(); if(entry.getValue() - 1 == 0) { // can't use: map.remove(entry.getKey()); // above wrong code would lead the iter can't find its next !!! iter.remove(); } else { map.put(entry.getKey(), entry.getValue() - 1); } } } else { map.put(num, 1); } } } // find the one that have highest value(count), that key should be the majority number int value = 0; int result = -1; for(Map.Entry<Integer, Integer> entry : map.entrySet()) { if(entry.getValue() > value) { value = entry.getValue(); result = entry.getKey(); } } return result; } }

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Data Structure is a way to organize data. It provides some methods to handle data stream, e.g. insert, delete, etc.

Linear Data Structure

Queue & Stack

Min Stack
Use two stacks, one is storing the input, when calling pop() or peek(), pop from another stack, which stores the minimum values from top to bottom.

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class MinStack {
    ArrayList<Integer> stack = new ArrayList<Integer>();
    ArrayList<Integer> minStack = new ArrayList<Integer>();
    public void push(int x) {
        stack.add(x);
        if(minStack.isEmpty() || minStack.get(minStack.size() - 1) >= x) {
            minStack.add(x);
        }
        return;
    }

    public void pop() {
        if(stack.isEmpty()) {
            return;
        }
        int elem = stack.remove(stack.size() - 1);
        if(!minStack.isEmpty() && minStack.get(minStack.size() - 1) == elem) {
            minStack.remove(minStack.size() - 1);
        }
        return;
    }

    public int top() {
        if(!stack.isEmpty()) {
            return stack.get(stack.size() - 1);
        }
        return 0;
    }

    public int getMin() {
        if(!minStack.isEmpty()) {
            return minStack.get(minStack.size() - 1);
        }
        return 0;
    }
}

Implement Queue by stacks
Use two stacks, one is for storing elements. When calling pop() or top(), pop the elements from the first stack and push them into the second one.

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public class Solution {
    private Stack<Integer> stack1;
    private Stack<Integer> stack2;

    public Solution() {
       // do initialization
       stack1 = new Stack<Integer>();
       stack2 = new Stack<Integer>();
    }
    
    public void push(int element) {
        stack1.push(element);
    }

    public int pop() {
        if(stack2.isEmpty()) {
            while(!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.pop();
    }

    public int top() {
        if(stack2.isEmpty()) {
            while(!stack1.isEmpty()) {
                stack2.push(stack1.pop());
            }
        }
        return stack2.peek();
    }
}

Largest Rectangle in Histogram
Brute force: totally O(n^2) windows and O(n) to find the minimun one in each window, so we need O(n^3) time.
Improve: for each number, search both side of it until find two smaller number, calculate the result, O(n^2) cost.
Best: use a stack to store the index (all increased heights), when faced with a smaller one, pop out from tha stack and calculate the area until the value bigger than the peek() one. This method cost O(n) time, with O(n) space in worst case (two passes).

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