LeetCode 17. Letter Combinations of a Phone Number

原题链接在这里：https://leetcode.com/problems/letter-combinations-of-a-phone-number/

题目：

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.

题解：

递归的stop condition 是index 达到 digits 的长度. e.g. "23", 开始index = 0, 加了一个字符后, index = 1, 再加一个 index = 2, 此时index == digits.length() 应该把sb加到res中，然后return.

每次加一个字符，这个字符数组是通过"23"中的对应数字，如2来确定的. n就是这个数字，可以用n来从keyBoard中找到对应的string.

Time Complexity: O(k^n), k是数字代表keyBoard中string的长度, n是digits的长度. Space(n).

AC Java:

public class Solution {
    public List<String> letterCombinations(String digits) {
        List<String> res = new ArrayList<String>();
        if(digits == null || digits.length() == 0){
            return res;
        }
        String [] keyBoard = {"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        dfs(digits, 0, new StringBuilder(), res, keyBoard);
        return res;
    }
    
    private void dfs(String digits, int index, StringBuilder sb, List<String> res, String [] keyBoard){
        if(index == digits.length()){
            res.add(sb.toString());
            return;
        }
        int n = digits.charAt(index) - '0'; //"23", index = 0, n = 2, 用来找对应keyBoard的string.
        for(int i = 0; i<keyBoard[n].length(); i++){
            sb.append(keyBoard[n].charAt(i));
            dfs(digits, index+1, sb, res, keyBoard);
            sb.deleteCharAt(sb.length()-1);
        }
    }
}

AC Python:

class Solution(object):
    def letterCombinations(self, digits):
        res = []
        if not digits:
            return res
        
        dic = {"2" : "abc", "3" : "def", "4" : "ghi", "5" : "jkl", "6" : "mno", "7" : "pqrs", "8" : "tuv", "9" : "wxyz"}
        self.dfs(digits, 0, "", dic, res)
        return res
        
    def dfs(self, digits, ind, item, dic, res):
        if(ind == len(digits)):
            res.append(item)
            return
        
        for c in dic[digits[ind]]:
            self.dfs(digits, ind + 1, item + c, dic, res)

类似Combinations, Generate Parentheses.