Leetcode 每日算法一題*****#17. 電話號碼的字母組合

Question：

給定一個僅包含數字 2-9 的字符串，返回全部它能表示的字母組合。

給出數字到字母的映射以下（與電話按鍵相同）。注意 1 不對應任何字母。

示例:

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

說明:
儘管上面的答案是按字典序排列的，可是你能夠任意選擇答案輸出的順序。

 

Answer:

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> tmp = new ArrayList<>();
        List<String> result = new ArrayList<>();
        Map map = new HashMap();
        map.put('2', "abc");
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");


        if (digits != null && digits.length() > 0) {

            char[] numbers = digits.toCharArray();

            for (char c : map.get(numbers[0]).toString().toCharArray()) {
                result.add(String.valueOf(c));
            }

            for (int i = 1; i < numbers.length; i++) {
                tmp.clear();
                tmp.addAll(result);
                result.clear();
                String shouldAdd = map.get(numbers[i]).toString();
                for (String s : tmp) {
                    for (char c1 : shouldAdd.toCharArray()) {
                        result.add(s + String.valueOf(c1));
                    }
                }

            }
        }
        return result;
    }
}

此用例運行花時5ms；順手貼一下運行耗時1ms大佬的解法；

class Solution {
    String[] codes = new String[]{"abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
    public List<String> letterCombinations(String digits) {
        if(digits == null || digits.equals("")){
            return new ArrayList<>();
        }
        List<String> result = new ArrayList<>();
        backtrace(digits, result, "", 1, digits.length());
        return result;
    }

    private void backtrace(String digits, List<String> list, String str, int deep, int n){
        if(deep == n){
            for(char c : codes[digits.charAt(deep-1) - '0' -2].toCharArray()){
                list.add(str+c);
            }
        }else{
            for(char c : codes[digits.charAt(deep-1) - '0' -2].toCharArray()){
                backtrace(digits, list, str+c, deep+1, n);
            }
        }
    }

}