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
python
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