LeetCode OJ:Letter Combinations of a Phone Number

Letter Combinations of a Phone Number

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.

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.

递归：

class Solution {
vector<string> result;
string str;
public:
void dfs(int k,string &digits,map<int,string> &mp){
if(k==digits.size()){
result.push_back(str);
return;
}
for(int i=0;i<mp[digits[k]-'0'].size();i++){
str+=mp[digits[k]-'0'][i];
dfs(k+1,digits,mp);
str.resize(k);
}

}
vector<string> letterCombinations(string digits) {
map<int,string> mp;
mp[2]="abc";mp[3]="def";
mp[4]="ghi";mp[5]="jkl";mp[6]="mno";
mp[7]="pqrs";mp[8]="tuv";mp[9]="wxyz";
dfs(0,digits,mp);
return result;
}
};

迭代：

class Solution {
public:
const vector<string> keyboard{" ","","abc","def",
"ghi","jkl","mno","pqrs","tuv","wxyz"};
vector<string> letterCombinations(string digits) {
vector<string> result(1,"");
for(auto d : digits){
const size_t n=result.size();
const size_t m=keyboard[d-'0'].size();

result.resize(n*m);
for(size_t i=0;i<m;++i)
copy(result.begin(),result.begin()+n,result.begin()+n*i);

for(size_t i=0;i<m;++i){
auto begin=result.begin();
for_each(begin+n*i,begin+n*(i+1),[&](string &s){
s+=keyboard[d-'0'][i];
});
}
}
return result;
}
};