python3

class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if not digits: #這句必不可少，處理空字串
            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

總結：

覺得比較精妙的幾句程式碼：

-if not x：（當x為真時不執行，x為假時，not x為真，才執行），與之同理的還有，if x is not None：（這種寫法最清楚，建議這麼寫）

• res = [i for i in p[digits[0]]] 一句話搞定，不需要再建立空列表，寫for迴圈，append值了，雖然意思一樣，但是程式碼簡化了不止一點點。
• res = [m+n for m in res for n in p[i]]，這句就更精妙了，兩重迴圈可以這麼寫，學習了！