[LeetCode] 17. Letter Combinations of a Phone Number ☆☆

 

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.

 

解法1: 

　　采用队列的方法，遍历digits的每一位数字，对于遍历到的数字，将队列中所有的字符串从头部移除，加上当前数字对应的字母后依次添加到队列后端。

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<>();
        
        if (digits == null || digits.length() == 0) {
            return res;
        }
        
        String[] letters = new String[]{"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        res.add("");
        
        for (int i = 0; i < digits.length(); i++) {
            int size = res.size();
            String str = letters[digits.charAt(i) - '2'];
            for (int j = 0; j < size; j++) {
                String front = res.remove(0);  // 不是remove(j)，每次都应该移除第一个字符串
                for (int k = 0; k < str.length(); k++) {
                    res.add(front + str.charAt(k));
                }
            }
        }
        return res;
    }
}

 

解法2: 

　　采用递归的方法，每次添加一个字母后进行下一次递归：

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<>();
        
        if (digits == null || digits.length() == 0) {
            return res;
        }
        
        String[] letters = new String[]{"abc", "def", "ghi", "jkl", "mno", "pqrs", "utv", "wxyz"};
        helper(res, letters, digits, "");
        return res;
    }
    
    public void helper(List<String> res, String[] letters, String digits, String temp) {
        if (digits.length() == 0) {
            res.add(temp);
            return;
        }
        String str = letters[digits.charAt(0) - '2'];
        for (int i = 0; i < str.length(); i++) {
            helper(res, letters, digits.substring(1), temp + str.charAt(i));
        }
    }
}

 

解法2: