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Given a digit string, return all possible letter combinations that the number could represent.
A mapping of digit to letters (just like on the telephone buttons) is given below.
Input:Digit string "23" Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].
这题是组合题，思路很简答，需要同时掌握迭代和递归两种方法。递归都很清楚，思路为dfs + backtracking,迭代版的思路为在之前生成的中间结果上再加一位构成现在的结果。
迭代版思路如下：
def letterCombinations(self, digits): """ :type digits: str :rtype: List[str] """ hash = {‘2‘:‘abc‘, ‘3‘:‘def‘,‘4‘:‘ghi‘,‘5‘:‘jkl‘,‘6‘:‘mno‘,‘7‘:‘pqrs‘,‘8‘:‘tuv‘,‘9‘:‘wxyz‘} #iterative method if not digits: return [] res = [[]] for i in digits: tmp = [] for j in hash[i]: for num in res: tmp.append(num + [j]) ＃取之前结果加1位 res = tmp + [] return map(lambda x: ‘‘.join(x),res)
递归版本：
class Solution(object): def letterCombinations(self, digits): """ :type digits: str :rtype: List[str] """ hash = {‘2‘:‘abc‘, ‘3‘:‘def‘,‘4‘:‘ghi‘,‘5‘:‘jkl‘,‘6‘:‘mno‘,‘7‘:‘pqrs‘,‘8‘:‘tuv‘,‘9‘:‘wxyz‘} comb = [] for i in digits: if i in hash: comb.append(hash[i]) if not comb: return [] n = len(comb) res = [] self.dfs(res, comb, [], 0, n) return res def dfs(self, res, comb, cur, index, n): if index == n: res.append(‘‘.join(cur)) return for j in xrange(len(comb[index])): cur.append(comb[index][j]) self.dfs(res, comb, cur, index+1, n) cur.pop()
总结而言python处理这种题不是很高效，主要在于string类型对象不可以修改。
Letter Combinations of a Phone Number
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原文地址：http://www.cnblogs.com/sherylwang/p/5764002.html
