LeetCode: "Letter Combinations of a Phone Number

一些小问题， “”的输出我的是[], 但是答案却是[""]， 比较难理解，3次过

class Solution {
public:
    void dfs(vector<string> &ret, string tmp, map<char, string> refer, string digits, int beg) {
        if (beg >= digits.size()) {
            ret.push_back(tmp);
            return;
        }
        if (digits[beg] <= '9' && digits[beg] >= '2') {
            for (int i = 0; i < refer[digits[beg]].size(); i++) {
                string r = refer[digits[beg]];
                dfs(ret, tmp+r[i], refer, digits, beg+1);
            }
        }
    }
    vector<string> letterCombinations(string digits) {
        // Start typing your C/C++ solution below
        // DO NOT write int main() function
        map<char, string> refer;
        refer['2'] = "abc";
        refer['3'] = "def";
        refer['4'] = "ghi";
        refer['5'] = "jkl";
        refer['6'] = "mno";
        refer['7'] = "pqrs";
        refer['8'] = "tuv";
        refer['9'] = "wxyz";
        vector<string> ret;
        string tmp = "";
        dfs(ret, tmp, refer, digits, 0);
        return ret;
    }
};

 加一个iterative的方法

class Solution {
public:
    vector<string> letterCombinations(string digits) {
        // IMPORTANT: Please reset any member data you declared, as
        // the same Solution instance will be reused for each test case.
        map<char, string> S;
        S['2'] = "abc", S['3'] = "def", S['4'] = "ghi", S['5'] = "jkl";
        S['6'] = "mno", S['7'] = "pqrs", S['8'] = "tuv", S['9'] = "wxyz";
        queue<string> que;
        vector<string> res;
        if (digits.size() == 0) {
            res.push_back("");
            return res;
        }
        for (int i = 0; i < S[digits[0]].size(); i++) que.push(string("")+S[digits[0]][i]);
        while (!que.empty()) {
            string tmp = que.front();
            que.pop();
            if (tmp.size() == digits.size()) res.push_back(tmp);
            else for (int i = 0; i < S[digits[tmp.size()]].size(); i++) que.push(tmp+S[digits[tmp.size()]][i]);
        }
        return res;
    }
};

 C#

public class Solution {
    public List<string> LetterCombinations(string digits) {
        Dictionary<char, string> S = new Dictionary<char, string>();
        S.Add('2', "abc"); S.Add('3', "def"); S.Add('4', "ghi"); S.Add('5', "jkl");
        S.Add('6', "mno"); S.Add('7', "pqrs"); S.Add('8', "tuv"); S.Add('9', "wxyz");
        Queue<string> que = new Queue<string>();
        List<string> ans = new List<string>();
        if (digits.Length == 0) return ans;
        foreach (var s in S[digits[0]]) que.Enqueue("" + (char)s);
        while (que.Count != 0) {
            string peek = que.Peek();
            que.Dequeue();
            if (peek.Length == digits.Length) ans.Add(peek);
            else foreach(var s in S[digits[peek.Length]]) que.Enqueue(peek + (char)s);
        }
        return ans;
    }
}