1,
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.
2, recursive
class Solution { public: vector<string> letterCombinations(string digits) { vector<string> res; string temp; rec(res, temp, digits, 0); return res; } private: void rec(vector<string> &res, string &temp, const string &digits, unsigned int level) { if(level == digits.length()) { res.push_back(temp); return; } string str = getMapChars(digits[level]); for(int i=0; i< str.length(); i++) { temp.push_back( str[i] ); rec(res, temp, digits, level+1); temp.pop_back(); } } string getMapChars(char digit) { if(digit<'2' || digit >'9') return ""; string arr[] = {"abc","def","ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; return arr[digit - '0' - 2]; } };
2. iterative java
public class Solution { public List<String> letterCombinations(String digits) { //init check List<String> res = new ArrayList<String>(); if(digits==null || digits.length()==0) return res; res.add(""); String[] map = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; for(int i=0; i<digits.length(); i++) { int index = digits.charAt(i) - '2' ; String s = map[index]; //check valid number int len = res.size(); for(int j=0; j<len; j++) { String str = res.get(j); res.set(j, str + s.charAt(0)); for(int m=1; m<s.length(); m++) { res.add(str + s.charAt(m)); } } } return res; } }
3, iterative c++
class Solution { public: vector<string> letterCombinations(string digits) { // Start typing your C/C++ solution below // DO NOT write int main() function const string letters[] = {"abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"}; vector<string> ret(1, ""); for (int i = 0; i < digits.size(); ++i) { for (int j = ret.size() - 1; j >= 0; --j) { const string &s = letters[digits[i] - '2']; for (int k = s.size() - 1; k >= 0; --k) { if (k) ret.push_back(ret[j] + s[k]); else ret[j] += s[k]; } } } return ret; } };