Saturday, October 6, 2018

오랫만에 풀어본 문제 serialized and desirialize

https://leetcode.com/problems/serialize-and-deserialize-binary-tree/description/ dfs 로 길을 갔었고 가는 곳 마다 흔적을 남겼고 다시 그 흔적을 따라 deserialized를 하면 풀리는 문제

Monday, September 17, 2018

sync와 async

123456789101112131415161718192021222324252627282930313233343536373839404142434445464748495051525354555657585960616263// sync
@Test
    public void zipTestSync(){

        long start = System.currentTimeMillis();
      // 비행편 가져오기
        Flight flight = lookupFlight("a");
      // 승객가져오기
        Passenger passenger = findPassanger("person");
      // 티켓 가져오기
        Ticket ticket = getTicket(flight, passenger);
      // 티켓 메일 보내기
        sendEmail(ticket);

        System.out.println(System.currentTimeMillis() - start);
    }

// async
@Test
    public void zipTestSingle(){
        long start = System.currentTimeMillis();

      // 비행기 lazyloading
        Single<Flight> flight = Single.defer(() -> Single.just(lookupFlight("a")))
                .subscribeOn(Schedulers.io());
      // 승객 lazyloading
        Single<Passenger> passenger = Single.defer(() ->Single.just(findPassanger("person")))
                .subscribeOn(Schedulers.io());
      // 비행기에서 승객을 zip으로 묶어서 다 로딩이된후 티켓을 가져오는데 .toBlocking으로 기다린다.
        flight.zipWith(passenger, (f, p) -> getTicket(f, p))
                .toBlocking()
                .value().getFlightName();
        System.out.println(System.currentTimeMillis()- start);

    }

 private Flight lookupFlight(String name) {
        try {
            Thread.currentThread().sleep(3000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("lokkupFlight : " + name);
        return Flight.builder().name(name).build();
    }

    private Passenger findPassanger(String name){
        try {
            Thread.sleep(4000);
        } catch (InterruptedException e) {
            e.printStackTrace();
        }
        System.out.println("findPassanger");
        return Passenger.builder().name(name).build();
    }

    private Ticket getTicket (Flight flight, Passenger passenger){
        return Ticket.builder().flightName(flight.name).passengerName(passenger.getName()).build();
    }

    private void sendEmail(Ticket ticket){
        System.out.println("sending ");
    }

Friday, August 10, 2018

인증샷이다

지금부터 8개월전 클라이밍을 시작했고 지금 와서 인증샷을 찍었다

[변화]
몸무게 78kg -> 76kg -2kg
근육 30.1kg -> 31kg +1kg
체지방 24.7 -> 21.1 -3.3kg
체지방률 31.5% -> 27.7%

종합 점수 : 64점 -> 70점

엄청난 변화인것 같다.

클라이밍을 1년정도는 더 할 생각이 있는데
그때는 75점은 넘지 않을까 기대되네.

Thursday, July 5, 2018

regular expression 사용 예제

오랫만에 패턴 매칭해서 결과값 구하는것 찾음.
regular expression은 내가 짠거는 쉬운데 남이 짠거는 잘 안읽힘. ㅠ

Pattern p = Pattern.compile("(([a-zA-Z][0-9][0-9]?)\\s*[\\/]\\s*.{3}\\s*[\\/]\\s*)");Matcher m = p.matcher(line); while(m.find())

T10 / name / AB12345

Sunday, March 18, 2018

line encoding

I solved this problem with for

somebody solved this one with reguar expression

Given a string, return its encoding defined as follows:

First, the string is divided into the least possible number of disjoint substrings consisting of identical characters
- for example, "aabbbc" is divided into ["aa", "bbb", "c"]
Next, each substring with length greater than one is replaced with a concatenation of its length and the repeating character
- for example, substring "bbb" is replaced by "3b"
Finally, all the new strings are concatenated together in the same order and a new string is returned.

Example

For s = "aabbbc", the output should be
lineEncoding(s) = "2a3bc".

Input/Output

[execution time limit] 3 seconds (java)
[input] string s

String consisting of lowercase English letters.

Guaranteed constraints:
4 ≤ s.length ≤ 15.
[output] string

Encoded version of s.

[Java] Syntax Tips

// Prints help message to the console
// Returns a string
// 
// Globals declared here will cause a compilation error,
// declare variables inside the function instead!
String helloWorld(String name) {
    System.out.println("This prints to the console when you Run Tests");
    return "Hello, " + name;
}

Sunday, March 4, 2018

codefights darkwilderness

My answer is O(n) and just use Math.abs

Given the positions of a white bishop and a black pawn on the standard chess board, determine whether the bishop can capture the pawn in one move.

The bishop has no restrictions in distance for each move, but is limited to diagonal movement. Check out the example below to see how it can move: