思路：用字典的形式保存号码的映射，实际组合是前一个数字串的组合加上后面一个数字的所有可能组合文章来源：https://www.toymoban.com/news/detail-685819.html
answer_dict={'2':['a','b','c'],'3':['d','e','f'],'4':['g','h','i'],'5':['j','k','l'],'6':['m','n','o'],'7':['p','q','r','s'],
'8':['t','u','v'],'9':['w','x','y','z']}
class Solution:
def letterCombinations(self, digits: str) -> list[str]:
digits=digits.replace('1','')
if not digits:
return []
digits_list=list(digits)
answer_list=answer_dict[digits_list[0]][:]
for digit in digits_list[1:]:
current_answer=[]
for each_answer in answer_list:
for each_char in answer_dict[digit]:
current_answer.append(each_answer+each_char)
answer_list=current_answer
print(answer_list)
return answer_list
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