Leetcode 17

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Question

Solution

考虑递归。每递归一次，digits都少一位，当digits为空的时候，递归返回一个结果。

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class Solution:
    def letterCombinations(self, digits: str) -> List[str]:
        if not digits:
            return []
        
        d = {'2':"abc",
             '3':"def",
             '4':"ghi",
             '5':"jkl",
             '6':"mno",
             '7':"pqrs",
             '8':"tuv",
             '9':"wxyz"
            }
        
        results = []
        self.dfs(d, digits, "", results)
        return results
    
    def dfs(self, d, digits, temp, results):
        if not digits:
            results.append(temp)
        else:
            for num in d[digits[0]]:
                self.dfs(d, digits[1:], temp+num, results)