#leetcode#Letter Combinations of a Phone Number

标签：leetcode

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"].

经典题目， iterative和recursive都要会写

iterative的是之前看的code ganker大神的解法，其实自己想也想得出来吧， 脑子太懒了

http://blog.csdn.net/linhuanmars/article/details/19743197

recursive的自己写了一下， 思路对了， 没能bug free

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<String>();
        if(digits == null || digits.length() == 0)
            return res;
            
        Map<Character, String> map = new HashMap<Character, String>();
        map.put('2', "abc");
        map.put('3', "def");
        map.put('4', "ghi");
        map.put('5', "jkl");
        map.put('6', "mno");
        map.put('7', "pqrs");
        map.put('8', "tuv");
        map.put('9', "wxyz");
        
        StringBuilder sb = new StringBuilder();
        
        helper(res, digits, map, sb, 0);
        
        return res;
    }
    
    private void helper(List<String> res, String digits, Map<Character, String> map, StringBuilder sb, int index){
        
        if(sb.length() == digits.length()){
            res.add(sb.toString());
            return;
        }
        
        String letters = map.get(digits.charAt(index));
        
        for(int i = 0; i < letters.length(); i++){
            sb.append(letters.charAt(i));
            helper(res, digits, map, sb, index + 1);
            sb.deleteCharAt(sb.length() - 1);
        }
    }
}

#leetcode#Letter Combinations of a Phone Number

标签：leetcode

原文地址：http://blog.csdn.net/chibaoneliuliuni/article/details/46754841