NO.17电话号码的字母组合

#dfs(深度优先遍历)
#将数字转为字典格式
#定义dfs函数(本质上回溯问题)，遍历dic[digit]返回值为dfs()函数，达到遍历的目的
#ps：由于存在空字符串的情况，必须设置加入条件！=''
def letterCombinations(self, digits: str) -> 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']} 
    res = []
    def dfs(digits,temp):
        if len(digits)==0:
            if temp !='':
                res.append(temp)
            return 
        digit = digits[0]
        for ch in dic[digit]:
            dfs(digits[1:],temp+ch)
    dfs(digits,'')
    return res

#笛卡尔积
#python人狂喜嗨还害
#本质上为
class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        conversion={'2':'abc','3':'def','4':'ghi','5':'jkl','6':'mno','7':'pqrs','8':'tuv','9':'wxyz'}
        if len(digits)==0:
            return [] 
        product=['']
        for k in digits:
            product=[i+j for i in product for j in conversion[k]] # for i for j 然后返回列表i与j的所有组合
        return product

# 探究i+j for i in a for j in b
a = ['a','b','c']
b = ['d','e']
res = [i+j for i in a for j in b]
res