[Leetcode] 17. Letter Combinations of a Phone Number

문제 바로가기

class Solution:
    def __init__(self):
        self.ans = list()
        self.digits = None
        self.digitLetterMapping = {
            '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'],
        }
        
    def letterCombinations(self, digits: str) -> List[str]:
        n = len(digits)
        if n == 0:
            return []
        
        self.digits = digits
        self.DFS(0, n, '')
        return self.ans
                
    def DFS(self, i, n, letterSoFar):
        if i == n:
            self.ans.append(letterSoFar)
            return
        
        for l in self.digitLetterMapping[self.digits[i]]:
            self.DFS(i + 1, n, letterSoFar + l)

내장함수 itertools를 사용한 방법

itertools.product(*iterables, repeat=1)
입력 이터러블들(iterables)의 데카르트 곱(Cartesian product, 곱집합).

import itertools
def letterCombinations(self, digits: str) -> List[str]:
    if len(digits) == 0:
        return []
    # Each digit map to an interator. (Recall strings are iterators)
    buttons = {'2':'abc','3':'def','4': 'ghi',\
               '5':'jlk','6':'mno','7':'pqrs',\
               '8':'tuv','9':'wxyz'}
    iterators = [buttons[d] for d in digits]
    # itertools.product(*args) returns the Cartesian product of the input iterables.
    return [''.join(x) for x in itertools.product(*iterators)]

참고

itertools.product() - 문서