leetcode17. Letter Combinations of a Phone Number

python3

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if not digits: #这句必不可少，处理空字符串
            return []
        p={}
        p['2'] = ['a','b','c']
        p['3'] = ['d','e','f']
        p['4'] = ['g','h','i']
        p['5'] = ['j','k','l']
        p['6'] = ['m','n','o']
        p['7'] = ['p','q','r','s']
        p['8'] = ['t','u','v']
        p['9'] = ['w','x','y','z']
        res = [i for i in p[digits[0]]]
        for i in digits[1:]:
            res = [m+n for m in res for n in p[i]]
        return res

总结：

觉得比较精妙的几句代码：

-if not x：（当x为真时不执行，x为假时，not x为真，才执行），与之同理的还有，if x is not None：（这种写法最清楚，建议这么写）

• res = [i for i in p[digits[0]]] 一句话搞定，不需要再创建空列表，写for循环，append值了，虽然意思一样，但是代码简化了不止一点点。
• res = [m+n for m in res for n in p[i]]，这句就更精妙了，两重循环可以这么写，学习了！