概述：
问题教程：HackerRank minion game practice tutorial
deck :: [Card]
deck = [Card rank suit | rank <- enumFrom minBound :: [Rank], suit <- enumFrom minBound :: [Suit]]
正如你所看到的，我是一个接一个，但我真的无法弄清楚错误的确切位置。
以下是代码：
Input: BAANANAS
Expected output: Kevin 19
My output: Kevin 18
任何注释掉的东西都可能用于调试（评论本身除外）
（注意：该函数在def minion_game(string):
# your code goes here
vowels = ('A', 'E', 'I', 'O', 'U')
def kevin(string):
kevin_list = []
for i in range(len(string)):
if string[i] in vowels:
return len(string) - i
#Find every possible combination beginning with that letter
for j in range(len(string)):
#Gets rid of white-space characters...for some reason....
if j >= i and string[i:j+1] not in kevin_list:
kevin_list.append(string[i:j+1])
return kevin_list
def stuart(string):
stuart_list = []
for i in range(len(string)):
if string[i] not in vowels:
#Find every possible combination beginning with that letter
for j in range(len(string)):
#Gets rid of white-space characters...for some reason....
if j >= i and string[i:j+1] not in stuart_list:
stuart_list.append(string[i:j+1])
return stuart_list
def points(words):
points_list = []
for substring in words:
points_list.append(string.count(substring))
return sum(points_list)
def calculateWinner(player1, score1, player2, score2):
if score1 > score2:
return '%s %d' %(player1, score1)
elif score2 > score1:
return '%s %d' %(player2, score2)
else:
return 'Draw'
#print(kevin(string))
#print("points:", points(kevin(string)))
print(calculateWinner("Stuart", points(stuart(string)), "Kevin", points(kevin(string))))
内调用，所以它被调用，不用担心。这只是它的定义）
答案 0 :(得分：0)
没关系。 如果重叠，BAANANAS中的.count（）方法不会将子字符串“ANA”计数两次。 如下所示： BA [ANA] NAS vs. BAAN [ANA] S. 它只计算其中一个，而不是两个。