Leetcode 17.电话号码的字母组合

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17.电话号码的字母组合

题目

给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。

给出数字到字母的映射如下（与电话按键相同）。注意 1 不对应任何字母。

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示例:
输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

说明:
尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。

方法1：回溯法

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class Solution:
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        phone = {'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 backtrack(combination, next_digits):
            # if there is no more digits to check
            if len(next_digits) == 0:
                # the combination is done
                output.append(combination)
            # if there are still digits to check
            else:
                # iterate over all letters which map
                # the next available digit
                for letter in phone[next_digits[0]]:
                    # append the current letter to the combination
                    # and proceed to the next digits
                    backtrack(combination + letter, next_digits[1:])

        output = []
        if digits:
            backtrack("", digits)
        return output
# test="23"
# print(Solution().letterCombinations(test))
# ['ad']
# ['ad', 'ae']
# ['ad', 'ae', 'af']
# --------------------
# ['ad', 'ae', 'af', 'bd']
# ['ad', 'ae', 'af', 'bd', 'be']
# ['ad', 'ae', 'af', 'bd', 'be', 'bf']
# --------------------
# ['ad', 'ae', 'af', 'bd', 'be', 'bf', 'cd']
# ['ad', 'ae', 'af', 'bd', 'be', 'bf', 'cd', 'ce']
# ['ad', 'ae', 'af', 'bd', 'be', 'bf', 'cd', 'ce', 'cf']
# --------------------
# --------------------
# ['ad', 'ae', 'af', 'bd', 'be', 'bf', 'cd', 'ce', 'cf']
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class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        phone = {'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 dfs(combination,index):
            if len(combination)==len(digits):
                output.append(combination)
                return

            for letter in phone[digits[index]]:
                dfs(combination+letter,index+1)

        output=[]
        if digits:
            dfs('',0)
        return output