题目

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。
给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。


示例：
输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

说明:
尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。

方案一

class Solution:
    def __init__(self):
        self.map = {
            "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 letterCombinations(self, digits: str) -> List[str]:
        if not digits:
            return []
        if len(digits) == 1:
            return self.map[digits]
        # 前 i 个 digits 组成的 ret
        res = self.letterCombinations(digits[:-1])
        ret = []
        for each in res:
            for word in self.map[digits[-1]]:
                ret.append(each + word)
        return ret

• 递归代码实现了回溯思想

  方案二（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

  原文

  https://leetcode-cn.com/explore/interview/card/2020-top-interview-questions/287/backtracking/1284/
  https://leetcode-cn.com/problems/letter-combinations-of-a-phone-number/solution/dian-hua-hao-ma-de-zi-mu-zu-he-by-leetcode/