Given a digit string, 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.

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

Note:
Although the above answer is in lexicographical order, your answer could be in any order you want.

方法一：標準正向遞歸，num == length時才append

class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        def helper(num, string):
            if (num == length):
                res.append(string)
            else:
                
 
for ls in dic[digits[num]]:
                    helper(num+1, string+ls)



        res = []
        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‘]
                }
        length = len(digits)
        if (length == 0):
            return []
        helper(0, ‘‘)
        return res
                      

方法二：temp臨時值傳值，python需要用res = copy.copy(tmp）

class Solution(object):
    ‘‘‘
    題意：輸出電話號碼對應的所有可能的字符串
    可以遞歸或直接模擬
    ‘‘‘
    def letterCombinations(self, digits):
        chr = ["", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"]
        res = []
        for i in range(0, len(digits)):
            num = int(digits[i])
            tmp = []
            for j in range(0, len(chr[num])):
                if len(res):
                    for k in range(0, len(res)):
                        tmp.append(res[k] + chr[num][j])
                else:
                    tmp.append(str(chr[num][j]))
            res = copy.copy(tmp)
        return res

Letter Combinations of a Phone Number