[LeetCode]题解（python）：017-Letter Combinations of a Phone Number

题目来源：

https://leetcode.com/problems/letter-combinations-of-a-phone-number/

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题意分析：

这道题是输入一段数字字符digits，在手机上每个数字所对应不同的字符。具体对应如图：

返回所有的数字字符对应的字符的可能。比如输入“123”，那么输出["*ad","*ae","*af","*bd","*be","*bf","*cd","*ce","cf"].

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题目思路：

看到这道题目让我想起了向量的笛卡尔乘积。这道题目可以用类似DP的方法去解决。dp[n] = dp[n -1]X最后一个字符对应的字符串，其中"X"代表内积，也就是说f("123") = f("1") X f("2") X f("3").首先建立一个字典，d = {'0':' ','1':'*','2':'abc','3':'def','4':'ghi','5':'jkl','6':'mno','7':'pqrs','8':'tuv','9':'wxyz'}；然后对利用笛卡尔乘积做法得到答案。

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代码（python）：

 class Solution(object):
     def addDigit(self,digit,ans):
         tmp = []
         for element in digit:
             if len(ans) == 0:
                 tmp.append(element)
             for s in ans:
                 tmp.append(s + element)
         return tmp
     def letterCombinations(self, digits):
         """
         :type digits: str
         :rtype: List[str]
         """
         ans = []
         d = {'':' ','':'*','':'abc','':'def','':'ghi','':'jkl','':'mno','':'pqrs','':'tuv','':'wxyz'}
         for element in digits:
             ans = self.addDigit(d[element],ans)
         return ans

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