题目描述：

方法一：回溯

class Solution:     def letterCombinations(self, digits):         """        :type digits: str        :rtype: List[str]        """         phone = {
   '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']}         def backtrack(combination, next_digits):             # if there is no more digits to check             if len(next_digits) == 0:                 # the combination is done                 output.append(combination)                 # if there are still digits to check             else:                     # iterate over all letters which map                     # the next available digit                 for letter in phone[next_digits[0]]:                     # append the current letter to the combination                     # and proceed to the next digits                     backtrack(combination + letter, next_digits[1:])         output = []         if digits: backtrack("", digits)         return output

二：

class Solution:     def letterCombinations(self, digits):         """        :type digits: str        :rtype: List[str]        """         m = {
   '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']}         if not digits:             return []         ls1 = ['']         for i in digits:             ls1 = [x + y for x in ls1 for y in m[i]]         return ls1