leetcode(17):Letter Combinations of a Phone Number

原题：

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”].

分析：求出由字母组合而成的集合，这些字母由给出的数字串中的单个数字来表示（如上图的电话按键）

思路是深度优先搜索，先取出一个数字，在深度遍历其余数字表示的字母，组成最后的字符串

public class Solution {
public static List<String> letterCombinations(String digits) {
List<String> result = new ArrayList<String>();
if(digits==null || digits.length()==0) return result;
Map<String,String[]> map = new HashMap<String,String[]>();
map.put("2", new String[]{"a","b","c"});
map.put("3", new String[]{"d","e","f"});
map.put("4", new String[]{"g","h","i"});
map.put("5", new String[]{"j","k","l"});
map.put("6", new String[]{"m","n","o"});
map.put("7", new String[]{"p","q","r","s"});
map.put("8", new String[]{"t","u","v"});
map.put("9", new String[]{"w","x","y","z"});
dfs(digits,0,"",map,result);//深度优先搜索
return result;
}
private static void dfs(String digits, int dx, String path,
Map<String, String[]> map, List<String> result) {
if(digits.length()==path.length()){//遍历到最后，将遍历的路径也就是得到的字母组合加入集合
result.add(path);
return ;
}
for (int i = dx; i < digits.length(); i++) {
for (String s : map.get(digits.substring(i,i+1))) {
dfs(digits,i+1,path+s,map,result);
}
}
}
}