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.
Although the above answer is in lexicographical order, your answer could be in any order you want.
void findpath(string &digits, int level, string &path, vector<string> &res, vector<string> &v){
if(level == digits.length()) {
res.push_back(path);
return;
}
if(digits[level] == 1 || digits[level] == 0) findpath(digits,level+1,path,res,v);
else{
for(int i = 0; i < v[digits[level]].length(); i++){
path.push_backdigits[level][i]);
findpath(digits,level+1,path,res,v);
path.pop_back();
}
}
}
vector<string> letterCombinations(string digits) {
vector<string> res;
string path;
vector<string> v(10);
v[1] = "abc";
v[2] = "def";
v[3] = "ghi";
v[4] = "jkl";
v[5] = "mno";
v[6] = "pqrs";
v[7] = "tuv";
v[8] = "wxyz";
findpath(digits, 0, path, res, v);
return res;
}
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"].
2->a,3->dNote:
Although the above answer is in lexicographical order, your answer could be in any order you want.
void findpath(string &digits, int level, string &path, vector<string> &res, vector<string> &v){
if(level == digits.length()) {
res.push_back(path);
return;
}
if(digits[level] == 1 || digits[level] == 0) findpath(digits,level+1,path,res,v);
else{
for(int i = 0; i < v[digits[level]].length(); i++){
path.push_backdigits[level][i]);
findpath(digits,level+1,path,res,v);
path.pop_back();
}
}
}
vector<string> letterCombinations(string digits) {
vector<string> res;
string path;
vector<string> v(10);
v[1] = "abc";
v[2] = "def";
v[3] = "ghi";
v[4] = "jkl";
v[5] = "mno";
v[6] = "pqrs";
v[7] = "tuv";
v[8] = "wxyz";
findpath(digits, 0, path, res, v);
return res;
}
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