Letter Combinations of a Phone Number

Given a digit string, 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.

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

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

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思路：典型backtracking，建立一个mapping的数组，然后进行NP搜索，注意最后递归结束条件是sb.length() == digits.length(), 因为，如果用start ==digits.length()，比如"22", 因为外层循环会首先从最后一个数字开始，最后一个数字的index也是digits.length()，这样会加入 "a" "b" "c" 到结果里面，所以这里的遍历必须以StringBuilder的长度为结束条件比较准确。

public class Solution {    public List<String> letterCombinations(String digits) {        List<String> list = new ArrayList<String>();        if(digits == null || digits.length() == 0) return list;        String[] mapArr = {"","","abc", "def", "ghi", "jkl","mno","pqrs","tuv","wxyz"};        StringBuilder sb = new StringBuilder();        collect(digits, sb, mapArr, 0, list);        return list;    }        public void collect(String digits, StringBuilder sb, String[] mapArr, int start, List<String> list) {        if( sb.length() == digits.length()){            list.add(sb.toString());            return;        }        for(int i=start; i<digits.length(); i++){            int index = digits.charAt(i)-'0';            String str = mapArr[index];            for(int j=0; j<str.length(); j++){                sb.append(str.charAt(j));                collect(digits, sb, mapArr, i+1, list);                sb.deleteCharAt(sb.length()-1);            }        }    }}