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January 6, 2018 16:48
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Programmation - 2013-2014-janvier - Solution - Ex8
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# Fonction | |
def amis(bdd,nom,d): | |
result = [] | |
if d == 0: | |
result.append(nom) | |
else: | |
for utilisateur in bdd[nom]: | |
result.append(utilisateur) | |
result += amis(bdd, utilisateur, d-1) | |
result_unique = set(result) # Permet de supprimer les doublons | |
return list(result_unique) | |
# Quelques tests (pas demandé dans l'énnoncé) | |
bddTest = {"Pierre" : ["Paul", "Luc", "Jacque"], "Jacque": ["Luc", "Paul"], "Adrien": ["Jacque"], "Arnaud" : [], "Paul": ["Luc"], "Luc": ["Adrien"]} | |
print("Adrien: (Résultat attendu: ['Jacque'])") | |
print(amis(bddTest, "Adrien", 1)) | |
print("") | |
print("Arnaud: (Résultat attendu: [])") | |
print(amis(bddTest, "Arnaud", 2)) | |
print("") | |
print("Pierre: (Résultat attendu: ['Paul', 'Luc', 'Jacque'])") | |
print(amis(bddTest, "Pierre", 1)) | |
print("") | |
print("Pierre: (Résultat attendu: ['Paul', 'Adrien', 'Luc', 'Jacque']) ") | |
print(amis(bddTest, "Pierre", 2)) | |
print("") | |
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