Given a digit string, 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.
Input:Digit string "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.
Solution 1:
private String[] btns={"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; public List<String> letterCombinations(String digits) { List<String> result = new ArrayList<>(); if(digits == null || digits.isEmpty()) { result.add(""); return result; } char[] letters = new char[digits.length()]; combine(result, letters, digits.toCharArray(), 0); return result; } public void combine(List<String> result, char[] letters, char[] digits, int start) { if(start == letters.length) { result.add(new String(letters)); return; } char[] chars = btns[digits[start]-'0'].toCharArray(); for(char c:chars) { letters[start] = c; combine(result, letters, digits, start+1); } }
Solution 2:
public List<String> letterCombinations(String digits) { List<String> result = new ArrayList<>(); result.add(""); if(digits == null || digits.isEmpty()) { return result; } String[] btns = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; for(int i=0; i<digits.length(); i++) { List<String> newResult = new ArrayList<>(); char[] chars = btns[digits.charAt(i)-'0'].toCharArray(); for(String s: result) { for(char c: chars) { newResult.add(s+c); } } result = newResult; } return result; }
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