신고 결과 받기
set을 사용해서 중복을 제거하고 시간 복잡도를 낮췄어야 했는데 그러지 못해서 좀 더 코드가 복잡해졌다.
내 풀이
def solution(id_list, report, k):
user_hash, count_hash = {}, {}
answer = []
for i in id_list :
user_hash[i] = []
count_hash[i] = 0
for r in report :
fromu, tou = r.split(" ")
if fromu not in user_hash[tou] :
user_hash[tou].append(fromu)
for i in id_list :
if len(user_hash[i]) >= k :
for u in user_hash[i] :
count_hash[u] += 1
for i in id_list :
answer.append(count_hash[i])
return answer다른 사람 풀이
def solution(id_list, reports, k):
stop = []
answer = [0] * len(id_list)
dicReports = {id: [] for id in id_list}
for i in set(reports):
report = i.split(' ')
stop.append(report[1])
dicReports[report[0]].append(report[1])
stop = set([i for i in stop if stop.count(i) >= k])
for key, value in dicReports.items():
for s in stop:
if s in value:
answer[id_list.index(key)] += 1
return answer