λ³Έλ¬Έ λ°”λ‘œκ°€κΈ°

πŸ€– μ½”λ”©ν…ŒμŠ€νŠΈ μ€€λΉ„/ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€

#ν”„λ‘œκ·Έλž˜λ¨ΈμŠ€ 파이썬 Lv.1 #μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜?

λ¬Έμ œμ„€λͺ…
: 길이가 n이고, "μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜...."와 같은 νŒ¨ν„΄μ„ μœ μ§€ν•˜λŠ” λ¬Έμžμ—΄μ„ λ¦¬ν„΄ν•˜λŠ” ν•¨μˆ˜, solution을 μ™„μ„±

λ¬Έμ œν’€μ΄

  1. iλ₯Ό 2둜 λ‚˜λˆˆ 값이 0이면 짝수 -> '수' μΆ”κ°€ 
  2. iλ₯Ό 2둜 λ‚˜λˆˆ 값이 0이 μ•„λ‹ˆλ©΄ ν™€μˆ˜ -> 'λ°•' μΆ”κ°€ 
  3. n = 3이면 i = 0, answer = '수' / i = 1, answer = 'μˆ˜λ°•' / i = 2, answer = 'μˆ˜λ°•μˆ˜' 
  4. n = 4이면 i = 0, answer = '수' / i = 1, answer = 'μˆ˜λ°•' / i = 2, answer = 'μˆ˜λ°•μˆ˜' / i = 3, answer = 'μˆ˜λ°•μˆ˜λ°•'

 

# solution

def solution(n):
    answer = ''
    for i in range(n):
        if i % 2 == 0:
            answer += '수'
        else:
            answer += 'λ°•'
    return answer

 

# λ‹€λ₯Έμ‚¬λžŒ 풀이

  1. 'μˆ˜λ°•' * n --> μž…λ ₯받은 n만큼 'μˆ˜λ°•' 반볡 
  2. solution(3)이면 'μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•' μ΄λ―€λ‘œ slicing ν•„μš” 
  3. λ”°λΌμ„œ μž…λ ₯받은 n만큼 좜λ ₯ν•΄μ£Όλ©΄ λœλ‹€. [:n]

 

def solution(n): 
	return ('μˆ˜λ°•'*n) [:n]

 

 

https://programmers.co.kr/learn/courses/30/lessons/12922

 

μ½”λ”©ν…ŒμŠ€νŠΈ μ—°μŠ΅ - μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜?

길이가 n이고, "μˆ˜λ°•μˆ˜λ°•μˆ˜λ°•μˆ˜...."와 같은 νŒ¨ν„΄μ„ μœ μ§€ν•˜λŠ” λ¬Έμžμ—΄μ„ λ¦¬ν„΄ν•˜λŠ” ν•¨μˆ˜, solution을 μ™„μ„±ν•˜μ„Έμš”. 예λ₯Όλ“€μ–΄ n이 4이면 "μˆ˜λ°•μˆ˜λ°•"을 λ¦¬ν„΄ν•˜κ³  3이라면 "μˆ˜λ°•μˆ˜"λ₯Ό λ¦¬ν„΄ν•˜λ©΄ λ©λ‹ˆλ‹€. μ œν•œ

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