17. Letter Combinations of a Phone Number 17.*的字母组合

Given a string containing digits from 2-9 inclusive, return all possible letter combinations that the number could represent.

A mapping of digit to letters (just like on the telephone buttons) is given below. Note that 1 does not map to any letters.

Example:

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

思路：知道要加字母，但是不知道具体的怎么加。

String candidate = candidates.get(sb.length());把一个字母里的元素拿出来是关键。当前长度，可以取出最后一位，好吧。

其实回溯法的backtrace函数里面不用写cc了，主函数写了就行了。backtrace函数里面只写退出函数就行了，这道题是sb的长度等于 

if (digits == null || digits.length() == 0) {
return new ArrayList();
}
cc直接返回一个新的就行了。先不用新建，新建会指定成string的类型。

sb的删除要用：deleteCharAt

backtrace里面也是sb，因为已经append了

Your input
"23"
stdout
candidates = [abc, def]
candidate = abc
  
candidate.charAt(i) = a
candidates = [abc, def]
candidate = def
  
candidate.charAt(i) = d
candidate.charAt(i) = e
candidate.charAt(i) = f
candidate.charAt(i) = b
candidates = [abc, def]
candidate = def
  
candidate.charAt(i) = d
candidate.charAt(i) = e
candidate.charAt(i) = f
candidate.charAt(i) = c
candidates = [abc, def]
candidate = def
  
candidate.charAt(i) = d
candidate.charAt(i) = e
candidate.charAt(i) = f

Output
["ad","ae","af","bd","be","bf","cd","ce","cf"]
Expected
["ad","ae","af","bd","be","bf","cd","ce","cf"]

class Solution {
    private String[] KEYS = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    
    public List<String> letterCombinations(String digits) {
        //cc
        List<String> candidates = new ArrayList<>();
        if (digits == null || digits.length() == 0) {
            return new ArrayList();
        }
        
        for (int i = 0; i < digits.length(); i++) {
            candidates.add(KEYS[digits.charAt(i) - '0']);
        }
        
        List<String> results = new ArrayList<>();
        backtrace(candidates, new StringBuilder(), results);
        
        return results;
    }
    
    public void backtrace(List<String> candidates, StringBuilder sb, List<String> results) {
        //exit
        if (sb.length() == candidates.size()) {
            results.add(sb.toString());
            return ;
        }
        
        String candidate = candidates.get(sb.length());
        for (int i = 0; i < candidate.length(); i++) {
            sb.append(candidate.charAt(i));
            backtrace(candidates, sb, results);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}