Java for Leetcode 017 letter combinations of a Phone number

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.

Problem Solving Ideas:

Using the DFS algorithm, the Java implementation is as follows:

    Static string[] Alpha = new string[] {    "", "1", "abc", "Def", "Ghi", "JKL", "MnO", "PQRS", "TUV", "WXYZ"    };    static StringBuilder sb = new StringBuilder ();      static  void Dfs (list<string> List, String digits, int cur) {        if (cur >= digits.length ())            List.add ( Sb.tostring ());        else {for            (int i = 0; i < Alpha[digits.charat (cur)-' 0 '].length (); i++) {                sb.append (Alpha[digits.charat (cur) -' 0 '].charat (i));                DFS (list, digits, cur + 1);                Sb.deletecharat (Sb.length ()-1);   }}} static public list<string> lettercombinations (String digits) {        list<string> List = new arraylist< String> ();        if (Digits.length () ==0)        return list;        DFS (list, digits, 0);        return list;    }

Idea two:

Wherever recursion is used, it can be solved using loops, so the Java implementation is as follows:

static public list<string> lettercombinations (String digits) {list<string> List = new arraylist<        String> ();        String[] Alpha = new string[] {"", "1", "abc", "Def", "Ghi", "JKL", "MnO", "PQRS", "TUV", "WXYZ"};        if (Digits.length () ==0) return list;        int[] Number = new Int[digits.length ()];//store each traversal character position int index = 0;            while (index>=0) {StringBuilder sb = new StringBuilder ();              for (int i=0; i<digits.length (); i++) Sb.append (Alpha[digits.charat (i)-' 0 '].charat (number[i]));              List.add (Sb.tostring ());              Each turn needs to be reset index to the end of index = Digits.length ()-1;                      while (index>=0) {if (Number[index] < (Alpha[digits.charat (index)-' 0 '].length ()-1)) {                      number[index]++;                  Break                      } else {Number[index] = 0;                  index--;  }            }} return list; }

Java for Leetcode 017 letter combinations of a Phone number