17.Letter Combinations of a Phone Number(Back-Track)

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"].

思路I：iteration，广度优先。三重for循环，分别循环数字字符串、每个数字对应的字母、已有的result。时间复杂度O(n*3*(3^n))

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        if(digits=="") return result;

        m.clear();
        m.insert(make_pair('',"abc"));
        m.insert(make_pair('',"def"));
        m.insert(make_pair('',"ghi"));
        m.insert(make_pair('',"jkl"));
        m.insert(make_pair('',"mno"));
        m.insert(make_pair('',"pqrs"));
        m.insert(make_pair('',"tuv"));
        m.insert(make_pair('',"wxyz"));

        result.push_back("");
        int resultSize;
        for(int i = ; i < digits.length(); i++){//traverse digits
            resultSize = result.size();
            cout << "size= " << resultSize << endl;
            for(int j = ; j < m[digits[i]].length(); j++){ //traverse the character in the digit(except first)
                for(int k = ; k < resultSize; k++){ //traverse all current result
                    result.push_back(result[k] + m[digits[i]][j]);
                }
            }
            for(int k = ; k < resultSize; k++){ //deal with first digit: directly add in the current result
                result[k] += m[digits[i]][];
            }
        }

        return result;
    }
private:
    unordered_map<char, string> m;
    vector<string> result;

};

思路II: recursion，深度优先。将最外层循环转换成递归，递归的depth表示是第几个数字。每次遍历要加上当前数字对应的字母（for循环），加好后递归调用。

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        m.clear();
        m.insert(make_pair('',"abc"));
        m.insert(make_pair('',"def"));
        m.insert(make_pair('',"ghi"));
        m.insert(make_pair('',"jkl"));
        m.insert(make_pair('',"mno"));
        m.insert(make_pair('',"pqrs"));
        m.insert(make_pair('',"tuv"));
        m.insert(make_pair('',"wxyz"));

        int len = digits.length();
        dfs("", digits,);
        return result;
    }
    void dfs(string s, string& digits, int depth){
        char c = digits[depth];

        for(int i = ; i < m[c].length(); i++){
            if(depth==digits.length()-) {
                result.push_back(s+m[c][i]);
            }
            else dfs(s+m[c][i], digits, depth+);
        }
    }
private:
    unordered_map<char, string> m; //map是用红黑树实现查找,O(logn); unordered_map是用Hash table实现查找,O(1)
    vector<string> result;
};