## Letter Combinations of a Phone Number

### 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"].

Note: Although the above answer is in lexicographical order, your answer could be in any order you want.

### 递归

// Letter Combinations of a Phone Number
// 时间复杂度O(3^n)，空间复杂度O(n)
public class Solution {
private static final String[] keyboard =
new String[]{ " ", "", "abc", "def", // '0','1','2',...
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

public List<String> letterCombinations(String digits) {
List<String> result = new ArrayList<>();
if (digits.isEmpty()) return result;
dfs(digits, 0, "", result);
return result;
}

private static void dfs(String digits, int cur, String path,
List<String> result) {
if (cur == digits.length()) {
return;
}
final String str = keyboard[digits.charAt(cur) - '0'];

for (char c : keyboard[digits.charAt(cur) - '0'].toCharArray()) {
dfs(digits, cur + 1, path + c, result);
}
}
}


### 迭代

// Letter Combinations of a Phone Number
// 时间复杂度O(3^n)，空间复杂度O(1)
public class Solution {
private static final String[] keyboard =
new String[]{ " ", "", "abc", "def", // '0','1','2',...
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

public List<String> letterCombinations(String digits) {
if (digits.isEmpty()) return new ArrayList<>();
List<String> result = new ArrayList<>();
for (char d : digits.toCharArray()) {
final int n = result.size();
final int m = keyboard[d - '0'].length();

// resize to n * m
for (int i = 1; i < m; ++i) {
for (int j = 0; j < n; ++j) {
}
}

for (int i = 0; i < result.size(); ++i) {
result.set(i, result.get(i) + keyboard[d - '0'].charAt(i/n));
}
}
return result;
}
}