Algorithm๐Ÿค/Python

[ํ”„๋กœ๊ทธ๋ž˜๋จธ์Šค] ์นด๋“œ ๋ญ‰์น˜ - for ๋ฆฌ์ŠคํŠธ, len, pop

ํŒŒ์นดํŒŒ์˜ค 2024. 6. 4. 11:50

 

# ๋ฌธ์ œ 

https://school.programmers.co.kr/learn/courses/30/lessons/159994

 

ํ”„๋กœ๊ทธ๋ž˜๋จธ์Šค

์ฝ”๋“œ ์ค‘์‹ฌ์˜ ๊ฐœ๋ฐœ์ž ์ฑ„์šฉ. ์Šคํƒ ๊ธฐ๋ฐ˜์˜ ํฌ์ง€์…˜ ๋งค์นญ. ํ”„๋กœ๊ทธ๋ž˜๋จธ์Šค์˜ ๊ฐœ๋ฐœ์ž ๋งž์ถคํ˜• ํ”„๋กœํ•„์„ ๋“ฑ๋กํ•˜๊ณ , ๋‚˜์™€ ๊ธฐ์ˆ  ๊ถํ•ฉ์ด ์ž˜ ๋งž๋Š” ๊ธฐ์—…๋“ค์„ ๋งค์นญ ๋ฐ›์œผ์„ธ์š”.

programmers.co.kr

 

 

# ๋ฌธ์ œ ์š”์•ฝ

 

 

# ํ•ต์‹ฌ ๊ฐœ๋…

 

1. for word in goal๋กœ ์ ‘๊ทผ

 

2. ๊ฐ๊ฐ์˜ ์นด๋“œ์— ๋Œ€ํ•ด์„œ ํฌ๊ธฐ๊ฐ€ 0์ดˆ๊ณผ์ธ์ง€ ํ™•์ธํ›„ ๋งž์„๋•Œ pop์‹œํ‚ดใ…

 

 

# ๋ฌธ์ œ ์ •๋‹ต

def solution(cards1, cards2, goal):
    for word in goal:
        if len(cards1) > 0 and word == cards1[0]:
            cards1.pop(0)
        elif len(cards2) > 0 and word == cards2[0]:
            cards2.pop(0)
        else:
            return "No"
    return "Yes"