LeetCode Letter Combinations of a Phone Number

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

与Combinations极其相似，也是DFS用递归来做。

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(k^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
.length(); i++){
sb.append(keyBoard
.charAt(i));
dfs(digits, index+1, sb, res, keyBoard);
sb.deleteCharAt(sb.length()-1);
}
}
}