我正在用給定的字符串制作代碼返回字符。我知道使用計數器功能更容易使用，但我盡量不使用它。這是我的代碼class Solution: def commonChars(self, A): dic = {} for i in range(len(A)): A[i] = list(A[i]) dic[i] = self.checkLetter(A[i]) print(dic) def checkLetter(self, Letter_list) : letter_cnt = {} for l in Letter_list: if l in letter_cnt : letter_cnt[l] += 1 else : letter_cnt[l] = 1 return letter_cnt我已經用字典制作了一個字母計數器，但我不知道下一步該做什么。你能給我一個提示嗎？# given input_1 : ["bella","label","roller"]# expected output is e,l,l because it is common in every string in the given list>>> ["e","l","l"]# result of my code>>> {0: {'a': 1, 'b': 1, 'e': 1, 'l': 2}, 1: {'a': 1, 'b': 1, 'e': 1, 'l': 2}, 2: {'r': 2, 'e': 1, 'l': 2, 'o': 1}}# given input_2 : ["cool","lock","cook"]# expected output>>> ["c","o"]
1 回答
慕哥9229398
TA貢獻1877條經驗 獲得超6個贊
reduce.only 添加一行的可能解決方案：
from functools import reduce
class Solution:
def commonChars(self, A):
dic = {}
for i in range(len(A)):
A[i] = list(A[i])
dic[i] = self.checkLetter(A[i])
print([char for char, count in reduce(lambda x, y:{k:min(x[k], y[k]) for k,v in x.items() if y.get(k)}, dic.values()).items() for _ in range(count)])
def checkLetter(self, Letter_list):
letter_cnt = {}
for l in Letter_list:
if l in letter_cnt:
letter_cnt[l] += 1
else:
letter_cnt[l] = 1
return letter_cnt
s = Solution()
s.commonChars(["bella","label","roller"])
# ['e', 'l', 'l']
s.commonChars(["cool","lock","cook"])
# ['c', 'o']
拆分它：
result_dict = reduce(lambda x, y:{k:min(x[k], y[k]) for k,v in x.items() if y.get(k)}, dic.values())
print([char for char, count in result_dict.items() for _ in range(count)])
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