17. 電話號碼的字母組合

暴力即可，深搜 or 叠代

class Solution {

    Map<Character, String[]> map = new HashMap<Character, String[]>();

    public List<String> letterCombinations(String digits) {
        if (null == digits || digits.length() == 0) return new LinkedList<>();

        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"});

        List<String> ans = new ArrayList<>();
        ans.add("");
        ans = dfs(ans, digits);
        ans.remove("");
        return ans;

    }

    public List<String> dfs(List<String> list, String digits) {
        if (digits.length() == 0) return list;
        char ch = digits.charAt(0);

        List<String> newList = new ArrayList<>();

        String[] strings = map.get(ch);
        for (String c : strings) {
            for (String s : list) {
                newList.add(s + c);
            }

        }

        return dfs(newList, digits.substring(1));
    }
}
 

17. 電話號碼的字母組合