LeetCode_17. Letter Combinations of a Phone Number 手机号码的字母组合，回溯法

题目：

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example:

Input: "23"
Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

本人代码：

class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        
        dic = {"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']}
        
        m = len(digits)
        
        result = []    #记录最终结果
        temp = []     #记录添加一组字母后的临时结果
        
        for i in range(m) :            
            if len(result) > 0 :
                for x in result :
                    for y in dic[digits[i]] :
                        temp.append(x+y)
            else :
                for y in dic[digits[i]] :
                    temp.append(y)

            result = temp
            temp = []

        return result

其他大牛算法：

1、典型的回溯法

 def letterCombinations(self, digits):
        mapping = {'2': 'abc', '3': 'def', '4': 'ghi', '5': 'jkl', 
                   '6': 'mno', '7': 'pqrs', '8': 'tuv', '9': 'wxyz'}
        if len(digits) == 0:
            return []
        if len(digits) == 1:
            return list(mapping[digits[0]])
        prev = self.letterCombinations(digits[:-1])
        additional = mapping[digits[-1]]
        return [s + c for s in prev for c in additional]

2、一行代码解决

  if '' == digits: return []
        kvmaps = {
            '2': 'abc',
            '3': 'def',
            '4': 'ghi',
            '5': 'jkl',
            '6': 'mno',
            '7': 'pqrs',
            '8': 'tuv',
            '9': 'wxyz'
        }
        return reduce(lambda acc, digit: [x + y for x in acc for y in kvmaps[digit]], digits, [''])