"Leetcode-Interview algorithm classic-java implementation" "017-letter combinations of a phone number (word combination on phone numbers)"

"017-letter combinations of a phone number (word combination on phone numbers)" "leetcode-Interview algorithm classic-java Implementation" "All topics Directory Index" Original Question

Given a digit string, return all possible letter combinations, the number could represent.
A mapping of Digit to letters (just as on the telephone buttons) is given below.

Input:"23"Output: ["ad""ae""af""bd""be""bf""cd""ce""cf"].

　　Note: Although the above answer is in lexicographical order, your answer could are in any order you want.

Main Topic

Given a string of numbers, returns all combinations of all characters on the number, as shown in the number-to-character mapping.
　　Note: Although the above results are arranged in character order, you can return the results in any order.

Thinking of solving problems

Use an array to hold the mapping of numbers and words, and according to the input of the number string, find the corresponding characters and combine the results.

Code Implementation

 Public classSolution {Privatestring[] Map = {"ABC","Def","Ghi","JKL","MnO","PQRS","TUV","WXYZ",    };Privatelist<string> result;//Store final results    Private Char[] chars;//Save the result of removing 0, 1 characters    Private Char[] curresult;//Store intermediate results    Private intEnd =0;The first unused position in the//character array    Private intHandle =0;//The number of characters that are currently being processed     PublicList<string>lettercombinations(String digits) {result =NewLinkedlist<> ();if(Digits! =NULL&& digits.length () >0) {chars = Digits.tochararray ();//string processing, minus 0 and 1            //Find the first 0 or 1 position             while(End < Digits.length () && chars[end]! =' 0 '&& Chars[end]! =' 1 ') {end++; } handle = end +1; while(Handle < Chars.length) {if(Chars[handle]! =' 0 '&& Chars[handle]! =' 1 ') {Chars[end] = Chars[handle];                end++;            } handle++; } Curresult =New Char[end];//While end, end is the length of a valid characterHandle =0;//point to the position of the first valid characterLettercombinations (); }returnResult }Private void lettercombinations() {if(Handle >= end) {Result.add (NewString (Curresult)); }Else{intnum = Chars[handle]-' 2 '; for(inti =0; I < map[num].length ();                i++) {Curresult[handle] = Map[num].charat (i);                handle++;                Lettercombinations ();            handle--; }        }    }}

Evaluation Results

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"Leetcode-Interview algorithm classic-java implementation" "017-letter combinations of a phone number (word combination on phone numbers)"