Given a string containing digits from 2-9inclusive, 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. Note that 1 does not map to any letters.
![[LeetCode] Letter Combinations of a Phone Number 电话号码的字母组合](https://img.zhankr.net/ifjfdokplab13137.png)
Example:
Input: "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.
这道题让我们求电话号码的字母组合，即数字2到9中每个数字可以代表若干个字母，然后给一串数字，求出所有可能的组合，相类似的题目有 Path Sum II，Subsets II，Permutations，Permutations II，Combinations，Combination Sum 和 Combination Sum II 等等。这里可以用递归 Recursion 来解，需要建立一个字典，用来保存每个数字所代表的字符串，然后还需要一个变量 level，记录当前生成的字符串的字符个数，实现套路和上述那些题十分类似。在递归函数中首先判断 level，如果跟 digits 中数字的个数相等了，将当前的组合加入结果 res 中，然后返回。我们通过 digits 中的数字到 dict 中取出字符串，然后遍历这个取出的字符串，将每个字符都加到当前的组合后面，并调用递归函数即可，参见代码如下：
解法一：
class Solution {
public:
vector<string> letterCombinations(string digits) {
if (digits.empty()) return {};
vector<string> res;
vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
letterCombinationsDFS(digits, dict, , "", res);
return res;
}
void letterCombinationsDFS(string& digits, vector<string>& dict, int level, string out, vector<string>& res) {
if (level == digits.size()) {res.push_back(out); return;}
string str = dict[digits[level] - ''];
for (int i = ; i < str.size(); ++i) {
letterCombinationsDFS(digits, dict, level + , out + str[i], res);
}
}
};
这道题也可以用迭代 Iterative 来解，在遍历 digits 中所有的数字时，先建立一个临时的字符串数组t，然后跟上面解法的操作一样，通过数字到 dict 中取出字符串 str，然后遍历取出字符串中的所有字符，再遍历当前结果 res 中的每一个字符串，将字符加到后面，并加入到临时字符串数组t中。取出的字符串 str 遍历完成后，将临时字符串数组赋值给结果 res，具体实现参见代码如下：
解法二：
class Solution {
public:
vector<string> letterCombinations(string digits) {
if (digits.empty()) return {};
vector<string> res{""};
vector<string> dict{"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
for (int i = ; i < digits.size(); ++i) {
vector<string> t;
string str = dict[digits[i] - ''];
for (int j = ; j < str.size(); ++j) {
for (string s : res) t.push_back(s + str[j]);
}
res = t;
}
return res;
}
};
Github 同步地址：
https://github.com/grandyang/leetcode/issues/17
类似题目：
参考资料：
https://leetcode.com/problems/letter-combinations-of-a-phone-number/
