翻译给定一个数字字符串，返回所有这些数字可以表示的字母组合。一个数字到字母的映射（就像电话按钮）如下图所示。输入：数字字符串“23”输出：[“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”]备注:尽管以上答案是无序的，如果你想的话你的答案可以是有序的。原图
原文Given a digit could represent.A mapping .Input:Digit string “23”Output: [“ad”, “ae”, “af”, “bd”, “be”, “bf”, “cd”, “ce”, “cf”].Note:Although lexicographical order, your answer could be in any order you want.代码
看样子我还是用C#顺手点，一气呵成……主要是用递归，因为不知道去判断。
public class Solution{IList<string> list = new List<string>();public Solution(){list.Insert(0, “abc”);list.Insert(1, “def”);list.Insert(2, “ghi”);list.Insert(3, “jkl”);list.Insert(4, “mno”);list.Insert(5, “pqrs”);list.Insert(6, “tuv”);list.Insert(7, “wxyz”);}public IList<string> LetterCombinations(string digits){IList<string> result = new List<string>();if (digits.Length == 0)return result;if (digits.Length == 1){foreach (var a in list.ElementAt(int.Parse(digits[0].ToString()) – 2)){result.Insert(0, a.ToString());}}int count = 0;IList<string> temp = LetterCombinations(digits.Substring(1, digits.Length – 1));foreach (var a in list.ElementAt(int.Parse(digits[0].ToString()) – 2)){foreach (var rest in temp){result.Insert(count++, a.ToString() + rest);}}return result;}}
以下是复制来的一段C++代码，，坦白地说，我写不出来这样的C++代码……
class Solution {public:vector<string> letterCombinations(string digits) {vector<string> ans;if(digits.size() == 0)return ans;int depth = digits.size();string tmp(depth, 0);dfs(tmp, 0, depth, ans, digits);return ans;}void dfs(string &tmp, int curdep, int depth, vector<string> &ans, string &digits){if(curdep >= depth){ans.push_back(tmp);return ;}for(int i = 0; i < dic[digits[curdep] – ‘0’].size(); ++ i){tmp[curdep] = dic[digits[curdep] – ‘0’][i];dfs(tmp, curdep + 1, depth, ans, digits);}return ;}private:string dic[10] = {{“”},{“”},{“abc”},{“def”},{“ghi”},{“jkl”},{“mno”},{“pqrs”},{“tuv”},{“wxyz”}};};
“人无完人金无足赤”，只要是人就不会是完美的，
