leetcode 140: Letter Combinations of a Phone Number

1,

Letter Combinations of a Phone Number

Total Accepted: 10538 Total Submissions: 40738

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;    }};