BiruLyu
6/5/2017 - 1:47 AM

17. Letter Combinations of a Phone Number(1st).java

"""
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.
"""
class Solution(object):
    def letterCombinations(self, digits):
        """
        :type digits: str
        :rtype: List[str]
        """
        if not digits:
            return [];
        DigitString = {
            '0' : " ",
            '1' : "*",
            '2' : "abc",
            '3' : "def",
            '4' : "ghi",
            '5' : "jkl",
            '6' : "mno",
            '7' : "pqrs",
            '8' : "tuv",
            '9' : "wxyz"
            }
            
        if digits in DigitString:
            return list(DigitString[digits]);
        else:
            return [a+b for a in self.letterCombinations(digits[0])
                        for b in self.letterCombinations(digits[1:])]
public class Solution {
    public List<String> letterCombinations(String digits) {
        HashMap<Character, String> keys = new HashMap<Character,String>();
        keys.put('2', "abc");
        keys.put('3', "def");
        keys.put('4', "ghi");
        keys.put('5', "jkl");
        keys.put('6', "mno");
        keys.put('7', "pqrs");
        keys.put('8', "tuv");
        keys.put('9', "wxyz");
        
        List<String> res = new ArrayList<String>();
        StringBuilder temp = new StringBuilder();
        if(digits == null || digits.length() == 0) return res;
        boolean[] flag = {true}; //set to false when meet invalid char
        backtracking(digits, keys, temp, res, 0, flag);
        return res;
        
    }
    
    private void backtracking(String digits, HashMap<Character, String> keys, StringBuilder temp, List<String> res, int start, boolean[] flag) {
        
        if(!flag[0]) return;
        int len = digits.length();
        if(start == len){
            res.add(temp.toString());
            return;
        }
        
        char num = digits.charAt(start);
        if(keys.containsKey(num)) {
            String candidates = keys.get(num);
            for(int j = 0; j < candidates.length(); j++) {
                temp.append(candidates.charAt(j));
                backtracking(digits, keys, temp, res, start + 1, flag);
                temp.setLength(temp.length() - 1);
            }
        } else {
            flag[0] = false;
        }

    }
}
public class Solution {
    String[] phoneButtons = new String[] {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};

    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<>();
        if (digits == null || digits.length() == 0) return res;
        helper(digits, 0, new StringBuilder(), res);
        return res;
    }
    
    public void helper(String digits, int index, StringBuilder sb, List<String> results) {
        if (index == digits.length()) {
            results.add(sb.toString());
            return;
        }
        int digit = digits.charAt(index) - '0';
        char[] characters = phoneButtons[digit].toCharArray();
        for (char c : characters) {
            sb.append(c);
            helper(digits, index+1, sb, results);
            sb.deleteCharAt(sb.length()-1);
        }
    }

    
}