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_answer
        print(answer_list)
        return answer_list