17. 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.

采用回溯的方法，遍历每一个数字对应的字母，相当于找到所有的排列组合，代码如下：

public class Solution {    String[] strs = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    public List<String> letterCombinations(String digits) {        List<String> res = new ArrayList<String>();        if (digits == null || digits.length() == 0) {            return res;        }        for(char ch: digits.toCharArray()) {            int digit = ch - '0';            if (digit < 2 || digit > 9) {                return res;            }        }        helper(digits, res, new StringBuilder(), 0);        return res;    }    private void helper(String digits, List<String> list, StringBuilder sb, int index) {        if (index == digits.length()) {            list.add(sb.toString());            return;        }        String chs = strs[digits.charAt(index) - '0'];        for (int i = 0; i < chs.length(); i ++) {            sb.append(chs.charAt(i));            helper(digits, list, sb, index + 1);            sb.deleteCharAt(sb.length() - 1);        }    }}

也可以采用FIFO的思路先进先出，每出来一个，加上下个数字的每一个字母，再存进去。代码如下：

public List<String> letterCombinations(String digits) {    LinkedList<String> ans = new LinkedList<String>();    String[] mapping = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};    ans.add("");    for(int i =0; i<digits.length();i++){        int x = Character.getNumericValue(digits.charAt(i));        while(ans.peek().length()==i){            String t = ans.remove();            for(char s : mapping[x].toCharArray())                ans.add(t+s);        }    }    return ans;}