https://programmers.co.kr/learn/courses/30/lessons/42840?language=python3
25분 소요
def solution(answers):
n = len(answers)
one = []
two = []
three = []
num1 = [1, 2, 3, 4, 5]
num2 = [2, 1, 2, 3, 2, 4, 2, 5]
num3 = [3, 3, 1, 1, 2, 2, 4, 4, 5, 5]
for i in range(n):
one.append(num1[i % 5])
two.append(num2[i % 8])
three.append(num3[i % 10])
ans = [0, 0, 0]
for i in range(n):
if answers[i] == one[i]:
ans[0] += 1
if answers[i] == two[i]:
ans[1] += 1
if answers[i] == three[i]:
ans[2] += 1
big = max(ans)
answer = []
for i in range(3):
if big == ans[i]:
answer.append(i+1)
return answer
def solution(answers):
pattern1 = [1,2,3,4,5]
pattern2 = [2,1,2,3,2,4,2,5]
pattern3 = [3,3,1,1,2,2,4,4,5,5]
score = [0, 0, 0]
result = []
for idx, answer in enumerate(answers):
if answer == pattern1[idx%len(pattern1)]:
score[0] += 1
if answer == pattern2[idx%len(pattern2)]:
score[1] += 1
if answer == pattern3[idx%len(pattern3)]:
score[2] += 1
for idx, s in enumerate(score):
if s == max(score):
result.append(idx+1)
return result
반복문 사용 시 몇 번째 반복문인지 확인이 필요할 수 있다. 이때, 인덱스 번호와 컬렉션의 원소를 tuple형태로 반환한다.
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