Given a string containing digits from 2-9 inclusive, 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.
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.
Solution:
Backtrace.
class Solution {
public:
vector<string> let{
"", "", "abc", "def", "ghi", "jkl","mno","pqrs","tuv","wxyz"
};
vector<string> ans;
vector<string> letterCombinations(string digits) {
if(digits.empty()){
return ans;
}
if(ans.empty()){
ans.push_back("");
}
vector<string> t;
for(char c: let[digits[0]-'0']){
for(string s: ans){
t.push_back(s + c);
}
}
ans = t;
return letterCombinations(digits.substr(1, digits.size()-1));
}
};
Iterative Solution:
Remember the index of each solution and add char accordingly (in groups):
len is the current group size, prev length is the previous group size
class Solution {
public:
vector<string> letterCombinations(string digits) {
if(digits.empty()){
return vector<string>();
}
int d[10] = {1,1,3,3,3,3,3,4,3,4};
string l[10] = {"","","abc","def","ghi","jkl","mno","pqrs","tuv","wxyz"};
int count = 1;
for(char c : digits){
count *= d[c-'0'];
}
int len = count;
int prev_len = count;
vector<string> ans(count,"");
for(char c: digits){
len/=d[c-'0'];
for(int i=0;i<count;i++){
ans[i]+=l[c-'0'][(i%prev_len)/len];
}
prev_len = len;
}
return ans;
}
};