目前，一些字符群组在伺机而动，准备battle一把，局面现在很焦灼。让我们来编写一个函数，按照评估规则，帮助吃瓜群众分辨下目前哪边更具有优势。返回胜方对象，如果是平局则返回“Tie!”表示平局。
评估规则：字母严格区分大小写，按照字母表出场顺序统计战力，A=1，B=2，···，Z=26，a=0.5, b = 1, ···, z=13。
示例：
输入： "One", "Two" ，输出："Two"
输入："ONE", "NEO" ，输出： "Tie!"
题目难度：简单
题目来源：CodeWars: Battle of the characters
def solution(x: str, y: str) -> str:
# your code here
assert solution("One", "Two") == "Two"
assert solution("One", "Neo") == "One"
assert solution("ONE", "NEO") == "Two"
题错了吧！assert solution(“ONE”, “NEO”) == “Two”
def solution(x: str, y: str) -> str:
x_value = sum([(26 - (ord('Z') - ord(char.upper()))) * (1 if char.isupper() else 0.5) for char in x])
y_value = sum([(26 - (ord('Z') - ord(char.upper()))) * (1 if char.isupper() else 0.5) for char in y])
if x_value > y_value:
return x
elif x_value < y_value:
return y
return "Tie!"
def solution(x: str, y: str) -> str:
sum_x = sum([ord(i) - 64 if i.isupper() else (ord(i) - 96) / 2 for i in x])
sum_y = sum([ord(i) - 64 if i.isupper() else (ord(i) - 96) / 2 for i in y])
if sum_x > sum_y:
return x
elif sum_x < sum_y:
return y
else:
return "Tie!"
assert solution("One", "Two") == "Two"
assert solution("One", "Neo") == "One"
assert solution("ONE", "NEO") == "Tie!"
def solution(x: str, y: str) -> str:
#ascii a-z:97-122 A-Z:65-90
"""
遍历字符串中的字符
判断字符是否大写
大写64-ord(i)
小写96-ord(i)
求和,比较
"""
xa,ya = 0,0
for i in x:
if i.isupper():
xa += 64 - ord(i)
else:
xa += (96 - ord(i))/2
for i in y:
if i.isupper():
ya += 64 - ord(i)
else:
ya += (96 - ord(i)) / 2
return y if xa > ya else (x if xa < ya else "Tie!")
assert solution("One", "Two") == "Two"
assert solution("One", "Neo") == "One"
assert solution("ONE", "NEO") == "Tie!"
def solution(a,b):
“”"
1、取出大写字母所代表的值
2、遍历字符串，把小写字母转化为大写后再除以2，
3、小写字母与大写字母进行相加
4、a 和 b 两值进行逻辑比较
:param a:
:param b:
:return:
“”"
def calc(str):
a_res = 0
for i in str:
if i.isupper():
a_res += ord(i)-64
else:
i = i.upper()
a_res += (ord(i)-64)/2
return a_res
if calc(a) > calc(b):
return a,calc(a),calc(b)
elif calc(a) == calc(b):
return "Tie"
else:
return b,calc(a),calc(b)
print(solution(“ONE”, “HELLO”))
def solution(x: str, y: str) -> str:
sum_x = sum([ord(i) - 64 if i.isupper() else (ord(i) - 96) / 2 for i in x])
sum_y = sum([ord(i) - 64 if i.isupper() else (ord(i) - 96) / 2 for i in y])
return x if sum_x > sum_y else (y if sum_x < sum_y else 'Tie!')
assert solution("One", "Two") == "Two"
assert solution("One", "Neo") == "One"
assert solution("ONE", "NEO") == "Tie!"
def solution(x: str, y: str) -> str:
dic = {chr(i + 65): i + 1 for i in range(26)}
for i in range(26):
dic[chr(i + 97)] = (i + 1) / 2
sum_x = sum([dic[x[i]] for i in range(len(x))])
sum_y = sum([dic[y[i]] for i in range(len(y))])
if sum_x > sum_y:
return x
elif sum_x < sum_y:
return y
else:
return "Tie!"
assert solution("One", "Two") == "Two"
assert solution("One", "Neo") == "One"
assert solution("ONE", "NEO") == "Tie!"