算法精粹(algorithm-essentials)

感谢soulmachine@github提供内容
## 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.

![Phone Keyboard](../images/phone-keyboard.png)

**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.


### 分析




### 递归

{% if book.java %}
```java
// 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 letterCombinations(String digits) {
List 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 result) {
if (cur == digits.length()) {
result.add(path);
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);
}
}
}
```
{% endif %}

{% if book.cpp %}
```cpp
// Letter Combinations of a Phone Number
// 时间复杂度O(3^n),空间复杂度O(n)
class Solution {
public:
const vector keyboard { " ", "", "abc", "def", // '0','1','2',...
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

vector letterCombinations (const string &digits) {
vector result;
if (digits.empty()) return result;
dfs(digits, 0, "", result);
return result;
}

void dfs(const string &digits, size_t cur, string path,
vector &result) {
if (cur == digits.size()) {
result.push_back(path);
return;
}
for (auto c : keyboard[digits[cur] - '0']) {
dfs(digits, cur + 1, path + c, result);
}
}
};
```
{% endif %}


### 迭代

{% if book.java %}
```java
// 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 letterCombinations(String digits) {
if (digits.isEmpty()) return new ArrayList<>();
List result = new ArrayList<>();
result.add("");
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) {
result.add(result.get(j));
}
}

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

{% if book.cpp %}
```cpp
// Letter Combinations of a Phone Number
// 时间复杂度O(3^n),空间复杂度O(1)
class Solution {
public:
const vector keyboard { " ", "", "abc", "def", // '0','1','2',...
"ghi", "jkl", "mno", "pqrs", "tuv", "wxyz" };

vector letterCombinations (const string &digits) {
if (digits.empty()) return vector();
vector result(1, "");
for (auto d : digits) {
const size_t n = result.size();
const size_t m = keyboard[d - '0'].size();

// resize to n * m
for (size_t i = 1; i < m; ++i) {
for (size_t j = 0; j < n; ++j) {
result.push_back(result[j]);
}
}

for (size_t i = 0; i < result.size(); ++i) {
result[i] = result[i] + keyboard[d - '0'][i/n];
}
}
return result;
}
};
```
{% endif %}