Leetcode17. Letter Combinations of a Phone Number

Leetcode17. Letter Combinations of a Phone Number

Given a string containing digits from 2-9 inclusive, 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.
Note that 1 does not map to any letters.

Example:

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

解法：
1.递归
2.利用队列

class Solution {
	//HashMap集合中key只能为引用数据类型，不能为基本类型
    Map<String, String> phone = new HashMap<String, String>() {{
	    put("2", "abc");
	    put("3", "def");
	    put("4", "ghi");
	    put("5", "jkl");
	    put("6", "mno");
	    put("7", "pqrs");
	    put("8", "tuv");
	    put("9", "wxyz");
  	}};
   	List<String> output = new ArrayList<String>();
   	////参数为目前已经产生的组合;接下来准备要输入的数字
  	public void backtrack(String combination, String next_digits) {
	    if (next_digits.length() == 0) {//先找递归出口
		    output.add(combination);
	    }else{
		    String digit = next_digits.substring(0, 1);//取next_digits的左边第一个字符
	        String letters = phone.get(digit);//get返回key对应的value
		    for (int i = 0; i < letters.length(); i++) {
			    String letter = phone.get(digit).substring(i, i + 1);//将value的字符依次取出
				backtrack(combination + letter, next_digits.substring(1));//前移一位
		    }
	    }
	}
    public List<String> letterCombinations(String digits) {
	    if (digits.length() != 0)
	      backtrack("", digits);
	    return output;
  	}
}

class Solution {
	public List<String> letterCombinations(String digits) {
		if(digits==null || digits.length()==0) {
			return new ArrayList<String>();
		}
		String[] letter_map = {//一个映射表，第二个位置是"abc“,第三个位置是"def"
			" ","*","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"
		};//这里也可以用map，用数组可以更节省点内存
		List<String> res = new ArrayList<>();
		res.add("");//先往队列中加入一个空字符
		for(int i=0;i<digits.length();i++) {			
			String letters = letter_map[digits.charAt(i)-'0'];//由当前遍历到的字符，取字典表中查找对应的字符串
			int size = res.size();			
			for(int j=0;j<size;j++) {//计算出队列长度后，将队列中的每个元素挨个拿出来				
				String tmp = res.remove(0);//每次都从队列中拿出第一个元素				
				for(int k=0;k<letters.length();k++) {//然后跟"def"这样的字符串拼接，并再次放到队列中
					res.add(tmp+letters.charAt(k));
				}
			}
		}
		return res;
	}
}

参考链接
leetcode题解