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-- 코드를 입력하세요

SELECT *

from (select T1.name, T1.datetime

from ANIMAL_INS T1

left join ANIMAL_OUTS T2

on T1.animal_id = T2.animal_id

where T2.animal_id is null

order by T1.datetime)

where rownum<=3;

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C#

 

 

 

using System;

public class Solution {
    public string solution(string s, string skip, int index) {
        string answer = "";
        
        for(int i = 0;i<s.Length;i++)
        {
            int asc = Convert.ToInt32(s[i]);
            
            for(int k = 0;k<index;k++)
            {
                asc++;
                if(asc > 122)
                {
                    asc = asc - 26;
                }
                
                for(int j = 0;j<skip.Length;j++)
                {
                    if(skip.Contains((char)asc))
                    {
                        asc++;
                        if(asc > 122)
                        {
                            asc = asc - 26;
                        }
                    }
                }
            }
        
            answer += (char)asc;
        }
        
        return answer;
    }
}

 

 

 

////////////////

 

 

 

자바

 

 

class Solution {
    public String solution(String s, String skip, int index) {
        StringBuilder answer = new StringBuilder();

        for (char letter : s.toCharArray()) 
        {
            char temp = letter;
            int idx = 0;
            
            while (idx < index) 
            {
                temp = temp == 'z' ? 'a' : (char) (temp + 1);
                
                if (!skip.contains(String.valueOf(temp))) 
                {
                    idx += 1;
                }
            }
            answer.append(temp);
        }

        return answer.toString();
    }
}

 

 

 

class Solution {
    public String solution(String s, String skip, int index) {
        String answer = "";

        for (char c : s.toCharArray()) 
        {
            for (int i = index; i > 0; i--) 
            {
                c++;
                if (c > 122) c -= 26;
                while (skip.contains(String.valueOf(c))) 
                {
                    c++;
                    if (c > 122) c -= 26;
                }
            }
            answer += c;
        }
        return answer;
    }
}

 

 

https://m.blog.naver.com/gboarder/90015619564

 

[C#] 문자의 ASCII코드 얻는 방법

Written by Youngil Kim, C#.NET Developer   ASCII코드란 숫자, 영소대문자등에 할당된...

blog.naver.com

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-- 코드를 입력하세요

SELECT

T1.ANIMAL_ID,

T1.NAME

from

ANIMAL_INS T1,

ANIMAL_OUTS T2

where T1.ANIMAL_ID = T2.ANIMAL_ID and

T1.DATETIME > T2.DATETIME

order by T1.DATETIME

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-- 코드를 입력하세요

SELECT DISTINCT T1.CART_ID

from CART_PRODUCTS T1,

CART_PRODUCTS T2

where

T1.CART_ID = T2.CART_ID and

T1.NAME = 'Yogurt' and

T2.NAME = 'Milk'

order by T1.CART_ID

/// 셀프 조인을 이용하는 방법도 있고

SELECT CART_ID

FROM CART_PRODUCTS

WHERE NAME = 'Yogurt'

INTERSECT

SELECT CART_ID

FROM CART_PRODUCTS

WHERE NAME = 'Milk'

ORDER BY CART_ID

/// INTERSECT를 이용하는 법도 있다.

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-- 코드를 입력하세요
SELECT NAME, DATETIME
FROM ANIMAL_INS
order by ANIMAL_ID DESC

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-- 코드를 입력하세요
SELECT ANIMAL_ID, NAME
from ANIMAL_INS
order by ANIMAL_ID

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-- 코드를 입력하세요

SELECT

NAME,

COUNT(NAME) as count

from ANIMAL_INS

group by NAME HAVING count(NAME) > 1

order by NAME;

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-- 코드를 입력하세요

SELECT ANIMAL_ID

from ANIMAL_INS

where NAME is not null

order by ANIMAL_ID ASC

 

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