首页 > 代码库 > Keypad Permutation

Keypad Permutation

2024-08-03 12:19:23   219人阅读

Problem

Phone has letters on the number keys. for example, number 2 has ABC on it, number 3 has DEF, 4 number has GHI,... , and number 9 has WXYZ. Write a program that will print out all of the possible combination of those letters depending on the input.  

 

Solution


HashMap

 1 public static ArrayList<String> letterCombinations(String digits) { 2   StringBuilder sb = new StringBuilder(); 3   ArrayList<String> res = new ArrayList<String>(); 4    5     if(digits == null) { 6         return res; 7     } 8      9     //hashmap to store the key pad info10     Map<Character, char[]> hm = new HashMap<Character, char[]>();11     hm.put(‘2‘, new char[]{‘a‘, ‘b‘, ‘c‘});12     hm.put(‘3‘, new char[]{‘d‘, ‘e‘, ‘f‘});13     hm.put(‘4‘, new char[]{‘g‘, ‘h‘, ‘i‘});14     hm.put(‘5‘, new char[]{‘j‘, ‘k‘, ‘l‘});15     hm.put(‘6‘, new char[]{‘m‘, ‘n‘, ‘o‘});16     hm.put(‘7‘, new char[]{‘p‘, ‘q‘, ‘r‘, ‘s‘});17     hm.put(‘8‘, new char[]{‘t‘, ‘u‘, ‘v‘});18     hm.put(‘9‘, new char[]{‘w‘, ‘x‘, ‘y‘, ‘z‘});19     20     helper(digits, res, sb, hm, 0);21     22     return res;23 24 }25 26 public static void helper(String digits, ArrayList<String> res, StringBuilder sb, Map<Character, char[]> hm, int pos) {27   if(pos == digits.length()) {28       res.add(sb.toString());29       return;30   }31   32   for(int i=0; i<hm.get(digits.charAt(pos)).length; i++) {33       sb.append(hm.get(digits.charAt(pos))[i]);34       helper(digits, res, sb, hm, pos+1);35       sb.deleteCharAt(sb.length()-1);36   }37 }

 

Keypad Permutation


联系
我们