在一些社交媒体App的博文中，会提供
【点赞】和
【点踩】的按钮，用来表达观众的态度。默认按钮的默认值都是空的。具体规则如下：点赞和点踩是互斥的，同一时间只能赞或者踩；连续点赞或者点踩按钮两次会在高亮和取消之间翻转。 请编写一个函数，接收用户的动作列表，返回最终的态度结果。如果是点赞则返回“赞”，如果是点踩则返回“踩”，如果既没有“赞”或者“踩”则返回无。
【示例】
输入：["踩", "赞"]
输出：赞
解释：用户首先点了踩，然后点了赞，最终这篇博文收到了赞。
题目难度：简单
题目来源：codewars-Likes Vs Dislikes
def solution(words:list)-> str:
# your code here
assert solution(["踩", "赞"]) == "赞"
assert solution(["赞", "赞"]) == "无"
assert solution(["踩"]) == "踩"
def solution(words: list) -> str:
# your code here
result = "无"
for i in words:
result = "无" if result == i else i
return result
assert solution(["踩", "赞"]) == "赞"
assert solution(["赞", "赞"]) == "无"
assert solution(["踩"]) == "踩"
def solution(words:list)-> str:
word = words[-1]
count = words.count(word)
if count % 2 == 0:
return "无"
return word
if __name__ == '__main__':
assert solution(["踩", "赞"]) == "赞"
assert solution(["赞", "赞"]) == "无"
assert solution(["踩"]) == "踩"
def solution(words:list)-> str:
tmp = "无"
for i in words:
tmp = "无" if i == tmp else i
return tmp
assert solution(["踩", "赞"]) == "赞"
assert solution(["赞", "赞"]) == "无"
assert solution(["踩"]) == "踩"
def solution(words:list)-> str:
return '无' if words[1:]==words[:-1] and len(words)>1 else words[-1]
def solution(words:list)-> str:
return '无' if len(set(words))==1 and len(words)>1 else words[-1]
def solution(words: list) -> str:
# your code here
length = len(words)
if length == 1:
return words[0]
elif length >= 2 and words[-1] != words[-2]:
return words[-1]
else:
return '无'
assert solution([“赞”, “赞”]) == “无”
这条case执行不同通过
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