Leetcode(Java)-17. 电话号码的字母组合

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。

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。

示例:

输入："23" 输出：["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"]. 说明: 尽管上面的答案是按字典序排列的，但是你可以任意选择答案输出的顺序。

class Solution {
    String[] dic = {"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    List<String> res = new ArrayList<>();
    public List<String> letterCombinations(String digits) {
        for(int i = 0;i < digits.length();i++)
        {
            int index = Integer.parseInt(String.valueOf(digits.charAt(i)))-2;
            if(res.size() == 0)
            {
                for(int j =0;j<dic[index].length();j++)
                {
                    res.add(String.valueOf(dic[index].charAt(j)));
                }
                continue;
            }
            List<String> temp = new ArrayList<>(res);
            res = new ArrayList<>();
            for(int j = 0; j <temp.size();j++)
            {
                String str = temp.get(j);
                for(int m=0;m < dic[index].length();m++)
                {
                    res.add(str+String.valueOf(dic[index].charAt(m)));
                }
            }
        }
        return res;
    }
}

回溯

class Solution {
    private String[] map = {"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    List<String> res = new ArrayList<String>();
    public List<String> letterCombinations(String digits) {
        if(digits == null || digits.length() ==0)
            return new ArrayList<String>();

        StringBuffer sb = new StringBuffer();
        solver(digits,0,sb);
        return res;
    }

    private void solver(String digits, int index, StringBuffer sb) {
        if(index == digits.length()){
            res.add(sb.toString());
            return;
        }

        String val = map[digits.charAt(index)-2];
        for(int i=0;i<val.length();i++)
        {
            sb.append(val.charAt(i));
            solver(digits,index+1,sb);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}