17. Letter Combinations of a Phone Number

题目描述

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"].

解题方法

a: def

b: def

c: def

/*Recursive way : 先找到數字字符對應的字串以做相對應的排列組合*/
    /*
    private static final String[] KEYS = { "", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

        public List<String> letterCombinations(String digits) {
            List<String> ret = new ArrayList<String>();
            if(digits == null || digits.length() == 0){
                return ret;
            }
            combination("", digits, 0, ret);
            return ret;
        }

    private void combination(String prefix, String digits, int offset, List<String> ret) {
            if (offset == digits.length()) {
                ret.add(prefix);
                return;
            }
            String letters = KEYS[(digits.charAt(offset) - '0')];
            for (int i = 0; i < letters.length(); i++) {
                combination(prefix + letters.charAt(i), digits, offset + 1, ret);
            }
    }

String[] key = new String[] {"0", "1", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
    public List<String> letterCombinations(String digits){
        LinkedList<String> lists = new LinkedList<>(); //* interface doesn't have peek() method
        if(digits == null || digits.length() == 0){
            return lists;
        }
        lists.add("");

        for(int i = 0; i < digits.length(); i++){
            char c = digits.charAt(i);
            String mapping = key[c - '0'];
            while( lists.peek().length() == i){
                String prefix = lists.remove();
                for(int j = 0; j < mapping.length(); j++){
                    lists.add(prefix + mapping.charAt(j));
                }
            }
        }
        return lists;
    }