Leetcode: 17. Letter Combinations of a Phone Number

Description

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.

Example

Input:Digit string "23"

Output: ["ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf"].

思路

思路1：

使用两个队列完成，代码如下:

思路2：

dfs

代码

class Solution {
public:
vector<string> letterCombinations(string digits) {
vector<string> res;
int len = digits.size();
if (len == 0) return res;

string letters[] = {
"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"
};
queue<string> Que, QueTemp;
int j = 0;
for (int i = 0; i < len; ++i){
if (digits[i] > '9' || digits[i] <= '1') return vector<string>();

j = digits[i] - '2';

if (Que.empty()){
for (int k = 0; k < letters[j].size(); ++k){
string tmp(charToString(letters[j][k]));
Que.push(tmp);
}
}
else{
while (!Que.empty()){
string tmp = Que.front();
Que.pop();
for (int k = 0; k < letters[j].size(); ++k){
QueTemp.push(tmp + letters[j][k]);
}
}
Que.swap(QueTemp);
}
}

while (!Que.empty()){
res.push_back(Que.front());
Que.pop();
}

return res;
}

string charToString(char c){
string res;
stringstream ss;
ss << c;
ss >> res;
return res;
}
};

class Solution {
public:
vector<string> letterCombinations(string digits) {
vector<string> res;
int len = digits.size();
if (len == 0) return res;
string letters[] = {
"", "", "abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"
};

string path;
dfs(res, digits, path, 0, letters);

return res;
}

void dfs(vector<string> &res, string& digits, string path, int index, string letters[]){
if (index == digits.size()){
res.push_back(path);
return;
}

for (int i = 0; i < letters[digits[index] - '0'].size(); ++i){
dfs(res, digits, path + letters[digits[index] - '0'][i], index + 1, letters);
}
}
};