Leetcode letter combinations of a phone number (alphanumeric combination of phone numbers)

translation

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。

Original

Original

Given a digitstring,returnAll possible letter combinations that  the  Numbercould represent. A Mapping ofDigit toLetters (Just Like on  theTelephone buttons) is given below. Input:digitstring "All"Output: ["AD","AE","AF","BD","Be","BF","CD","CE","CF"]. Note:although the aboveAnswer is inchlexicographical order, your answer could beinchAny order you want.

Code

It seems that I still use C # handy point, one go ... Mostly with recursion, because I don't know < Span class= "Mrow" id= "mathjax-span-67" > d i g i t s The length of the end is how much, impossible to write countless < Span class= "Mrow" id= "mathjax-span-75" > f o r e a c h To judge.

 Public classsolution{ilist<string> list =Newlist<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"); } Publicilist<string>lettercombinations(stringdigits) {ilist<string> result =Newlist<string> ();if(digits. Length = =0)returnResultif(digits. Length = =1)        {foreach(varAinchList. ElementAt (int. Parse (digits[0]. ToString ())-2) {result. Insert (0, a.tostring ()); }        }intCount =0; ilist<string> temp = lettercombinations (digits. Substring (1, digits. Length-1));foreach(varAinchList. ElementAt (int. Parse (digits[0]. ToString ())-2))        {foreach(varRestinchTemp) {result.            Insert (count++, a.tostring () + rest); }        }returnResult }}

Here is a copy of the C + + code, frankly, I can not write such C + + code ...

classSolution { Public: vector<string>Lettercombinations (stringDigits) { vector<string>Ansif(digits.size () = =0)returnAnsintdepth = Digits.size ();stringTMP (Depth,0); DFS (TMP,0, depth, ans, digits);returnAns }voidDfsstring&tmp,intCURDEP,intDepth vector<string>&ans,string&digits) {if(CURDEP >= depth) {ans.push_back (TMP);return; } for(inti =0; I < DIC[DIGITS[CURDEP]-' 0 '].size (); + + i) {TMP[CURDEP] = Dic[digits[curdep]-' 0 '][i]; DFS (TMP, CURDEP +1, depth, ans, digits); }return; }Private:stringdic[Ten] = {{""},{""},{"ABC"},{"Def"},{"Ghi"},{"JKL"},{"MnO"},{"PQRS"},{"TUV"},{"WXYZ"}};};

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Leetcode letter combinations of a phone number (alphanumeric combination of phone numbers)