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”].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);
}
}
}
}