17-电话号码的组合

        给定一个仅包含数字 2-9 的字符串，返回所有它能表示的字母组合。答案可以按 任意顺序 返回。给出数字到字母的映射如下（与电话按键相同）。注意1不对应任何字母。输入：digits = "23"；输出：["ad","ae","af","bd","be","bf","cd","ce","cf"]

        要求解字母组合，则是从每一个按键中找到选出一个元素，如果要求输出的字母数量就很好求，但这里要的是输出字母，所以比较麻烦。假设入如为23两个按键，输出字母则为从2中选一个字符，从3中再选一个字符。一般求解组合之类的问题，第一感觉都是回溯法。

package leecodeOff;

import java.util.ArrayList;
import java.util.HashMap;
import java.util.List;
import java.util.Map;


/*
 * 17.电话号码的字母组合
 */
public class Demo17 {
	

	public static void main(String args[]) {
		Demo17 demo = new Demo17();
		String digits = "23";
		List<String> list = demo.letterCombinations(digits);
		for (int i = 0; i < list.size(); i++) {
			System.out.println(list.get(i));
		}
	}
    
	
	char[] char_2 = { 'a', 'b', 'c' };
	char[] char_3 = { 'd', 'e', 'f' };
	char[] char_4 = { 'g', 'h', 'i' };
	char[] char_5 = { 'j', 'k', 'l' };
	char[] char_6 = { 'm', 'n', 'o' };
	char[] char_7 = { 'p', 'q', 'r', 's' };
	char[] char_8 = { 't', 'u', 'v' };
	char[] char_9 = { 'w', 'x', 'y', 'z' };
	
	public List<String> letterCombinations(String digits) {
		List<String> letterCombinations = new ArrayList<String>();
		StringBuffer letterCombination = new StringBuffer();
		if(digits.length() == 0) {
			return letterCombinations;
		}
		Map<Character, char[]> map = getMap();
		backTrack(letterCombinations, 0, digits, map, letterCombination);
		return letterCombinations;
	}

	public void backTrack(List<String> letterCombinations, int index, String digits, Map<Character, char[]> map,
			StringBuffer letterCombination) {
		if (index == digits.length()) {
			letterCombinations.add(letterCombination.toString());
		} else {
			char temp = digits.charAt(index);
			char[] tempArray = map.get(temp);
			for (int i = 0; i < tempArray.length; i++) {
				letterCombination.append(tempArray[i]);
				backTrack(letterCombinations, index + 1, digits, map, letterCombination);
				letterCombination.deleteCharAt(index);
			}
		}
	}

	Map<Character, char[]> map = null;

	public Map<Character, char[]> getMap() {
		if (map != null) {
			return map;
		} else {
			map = new HashMap<Character, char[]>();
			map.put('2', char_2);
			map.put('3', char_3);
			map.put('4', char_4);
			map.put('5', char_5);
			map.put('6', char_6);
			map.put('7', char_7);
			map.put('8', char_8);
			map.put('9', char_9);
			return map;
		}
	}

}