leetcode-mid-backtracking-17. Letter Combinations of a Phone Number

mycode  68.26%

class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """           
        if digits == "":
            return []
        dict = {'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 dfs(index,temp):
            if index==self.len:
                self.res.append(temp)
                return 
            num = digits[index]
            s = dict[num]
            for i in s:
                dfs(index+1,temp+i)   
        self.res = []
        self.len = len(digits)
        dfs(0,"")
        return self.res

参考：

def letterCombinations(digits):
        def dfs(num, string):
            if num == length:
                res.append(string)
                return
            for letter in dict[digits[num]]:
                    dfs(num+1, string+letter)
        
        dict = {'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']
                }
        res = []
        length = len(digits)
        dfs(0, '')
        return res