leetcode-17-电话号码的字母组合’

题目描述：

方法一：回溯

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        phone = {'': ['a', 'b', 'c'], '': ['d', 'e', 'f'], '': ['g', 'h', 'i'], '': ['j', 'k', 'l'], '': ['m', 'n', 'o'], '': ['p', 'q', 'r', 's'], '': ['t', 'u', 'v'], '': ['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 = {'': ['a', 'b', 'c'], '': ['d', 'e', 'f'], '': ['g', 'h', 'i'], '': ['j', 'k', 'l'], '': ['m', 'n', 'o'], '': ['p', 'q', 'r', 's'], '': ['t', 'u', 'v'], '': ['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