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"].
思路：假如字符串长度已知，直接采用多重循环效率更高一些。不过这道题长度未知，使用递归，一层一层地递归下去，这种思路很重要，最后需要注意得是检查特殊情况null和""。
public List<String> letterCombinations(String digits) {
List<String> list=new ArrayList<String>();
if(digits==null || digits.length()==0)
return list;
digits=digits.replace("1","");
Map<Character, String> map=new HashMap<Character,String>();
map.put(‘1‘, "");
map.put(‘2‘, "abc");
map.put(‘3‘, "def");
map.put(‘4‘, "ghi");
map.put(‘5‘, "jkl");
map.put(‘6‘, "mno");
map.put(‘7‘, "pqrs");
map.put(‘8‘, "tuv");
map.put(‘9‘, "wxyz");
map.put(‘0‘, " ");
recurseStr(map , digits , 0 , list , "");
return list;
}
private void recurseStr(Map<Character,String> map,String digits,int depth,List<String> list,String re)
{
if(depth==digits.length())
{
list.add(re);
return;
}
String temp=map.get(digits.charAt(depth));
for(int i=0;i<temp.length();i++)
{
recurseStr(map,digits,depth+1, list, re+temp.charAt(i));
}
}
[LeetCode] Letter Combinations of a Phone Number
原文：http://www.cnblogs.com/maydow/p/4667687.html