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.
思路: 深度优先, 回溯
public class Solution { public List<String> letterCombinations(String digits) { List<String> result = new ArrayList<String>(); if(digits == null || digits.length() == 0) { return result; } Map<Character,char[]> map = new HashMap<Character, char[]>(); map.put('1', new char[]{}); map.put('2', new char[]{'a', 'b', 'c'}); map.put('3', new char[]{'d', 'e', 'f'}); map.put('4', new char[]{'g', 'h' , 'i'}); map.put('5', new char[] { 'j', 'k', 'l' }); map.put('6', new char[] { 'm', 'n', 'o' }); map.put('7', new char[] { 'p', 'q', 'r', 's' }); map.put('8', new char[] { 't', 'u', 'v'}); map.put('9', new char[] { 'w', 'x', 'y', 'z' }); StringBuilder sb = new StringBuilder(); helper(result, digits, map, sb); return result; } public void helper(List<String> result, String digits, Map<Character, char[]> map, StringBuilder sb) { if(sb.length() == digits.length()) { result.add(sb.toString()); return; } for(char c : map.get(digits.charAt(sb.length()))) { sb.append(c); helper(result, digits, map, sb); sb.deleteCharAt(sb.length() - 1); } } }