leetcod17：电话号码的字母组合

**思路：和产生集合的所有子集方法类似。
例：‘23’
第一轮：
一：[‘a’,‘b’,‘c’]
遍历第一轮生成的集合，生成第二轮
二：[“ad”,“bd”,“cd”]+[“ae”,“be”,“ce”]+[“af”,“bf”,“cf”]

代码

   #写法一
    class Solution:
        def letterCombinations(self, digits: str) -> List[str]:
            if not digits:
                return []
            b = ['','','abc','def','ghi','jkl','mno','pqrs','tuv','wxyz']   #加两个空字符串是为了查找对应字符更方便(学到了)
            res = ['']
            for i in digits:
                tmp = []     ###每次更新tmp这个临时数组
                for j in b[int(i)]:
                    for z in res:
                        tmp.append((z+j))
                res = tmp

    
    #写法二
    class Solution:
        def letterCombinations(self, digits: str) -> List[str]:
            result = []
            maps = {'2':'abc','3':'def','4':'ghi','5':'jkl','6':'mno','7':'pqrs','8':'tuv','9':'wxyz'}
            for digit in digits:
                temp = []
                for char in maps[digit]:
                    if len(result) == 0:
                        temp += [char]
                    else:
                        temp += [s+char for s in result]
                result = temp
            return result