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

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_answerprint(answer_list)return answer_list