Monogram phone number. (A day, to prevent dementia)

Given a string contains only numeric characters 2-9, it can return all letter combinations indicated.

Given digital map to letters as follows (the same telephone key). Note 1 does not correspond to any alphabet.

Example:

输入："23"
输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

Source: stay button (LeetCode)
link: https: //leetcode-cn.com/problems/letter-combinations-of-a-phone-number
copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.

1  determined to contain only numbers 2-9 string, it returns all letter combinations indicated.
2  
. 3 gives the following mapping numbers to letters (the same telephone key). Note 1 does not correspond to any alphabet.
. 4  
. 5  
. 6  
. 7   
. 8  
. 9  Example:
 10  
. 11 Input: "23"
 12 Output: [ "ad", "ae ", "af", "bd", "be", "bf", "cd", "ce" , "cf" ].
 13  
14  source: stay button (leetCode)
 15 links: HTTPS: // leetcode-CN. 
COM / Problems / Combinations-of-Letter-A-Phone-Number 16  copyrighted by deduction from all networks. Commercial reprint please contact the authorized official, non-commercial reprint please indicate the source.
. 17  
18 is  
. 19  
20 is  public  class Leetcode17 {
 21 is      //
        
       List<String> output = new ArrayList<String>();
34 public static void main(String[] args) {
35     List<String> strings =new Leetcode17().letterCombinations("23");
36     System.out.println(strings);
37 }
38 //digits用于存放输入的数字
39 public List<String> letterCombinations(String digits) {
40     if (digits.length() != 0)
41       backtrack("", digits);
42     return output;
43   }
44  // recursive Combination represented temporary strings have been combined, next_digits represents the remaining figures, 
45  public  void BackTrack (Combination String, String next_digits) {
 46 is      // recursive method outlet, which represents a combination splicing is completed, deposited into the result set. 
47      IF (next_digits.length () == 0 ) {
 48          output.add (Combination);
 49      } the else {
 50          // number remaining not taken in the first traverse 
51 is          String digit for next_digits.substring = (0,1 );
 52          // acquired digital corresponding to all of the characters 
53 is          String Letters = phone.get (digit for);
 54 is          //Traversing a character, length () is a method of String, String This method string into a character array display its length. 
55          for ( int I = 0; I <letters.length (); I ++ ) {
 56 is              // traversing a character 
57 is               String Letter = phone.get (digit for) .substring (I, I +. 1 );
 58               // recursive call next_digits .substring (1) represents all numbers from first to last place. 
59               BackTrack (+ Combination Letter, next_digits.substring (. 1 ));
 60              
61 is          }
 62 is      }
 63 is      
64  }
 65  }
 66

输入：23
输出：[ad, ae, af, bd, be, bf, cd, ce, cf]
过程是：先取得"23"的第一位数字"2",然后取得对应的字符"abc"，然后遍历"abc"，继续调用递归函数 backtrack("a","23".substring(1))，所以参数还剩"3"，
取得"3"对应的字符"def"，然后遍历，在调用递归函数 backtrack("a"+"d","3".substring(1))，此时已经没有剩余的数字，就到了递归的的出口，把"ad"放入
结果集，然后回到上一层的for循环，递归调用 backtrack("a"+"e","3".substring(1))。。。。。

Monogram phone number. (A day, to prevent dementia)

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