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.

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Example:

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

Although the above answer is in lexicographical order, your answer could be in any order you want.

难度:medium

题目:
给定字符串包含数字2到9,返回所有字母组合。数字字母的对应关系如下图所示。注意1不与任何字母对应。

思路:
组合,递归

Runtime: 2 ms, faster than 80.94% of Java online submissions for Letter Combinations of a Phone Number.
Memory Usage: 26.4 MB, less than 14.87% of Java online submissions for Letter Combinations of a Phone Number.

class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> result = new ArrayList();
        if (digits.isEmpty()) {
            return result;
        }
        
        Map<Character, String> mcs = new HashMap<>();
        mcs.put('2', "abc");
        mcs.put('3', "def");
        mcs.put('4', "ghi");
        mcs.put('5', "jkl");
        mcs.put('6', "mno");
        mcs.put('7', "pqrs");
        mcs.put('8', "tuv");
        mcs.put('9', "wxyz");
        
        StringBuilder s = new StringBuilder();
        letterCombinations(digits, 0, mcs, result, s);

        return result;
    }
    
    private void letterCombinations(String digits, int digitIdx, Map<Character, String> mcs, List<String> result, StringBuilder s) {
        if (s.length() == digits.length()) {
            result.add(s.toString());
            return;
        }
        char[] cs = digits.toCharArray();
        for (int i = digitIdx; i < cs.length; i++) {
            String str = mcs.get(cs[i]);
            for (int j = 0; j < str.length(); j++) {
                s.append(str.charAt(j));
                letterCombinations(digits, i + 1, mcs, result, s);
                s.deleteCharAt(s.length() - 1);
            }
        }
    }
}

linm
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