LeetCode 17. Letter Combinations of a Phone Number (迭代)

迭代和递归一直都是我的弱项（说的好像我有强项一样- - ）  不过迭代和递归感觉很多用在无法确定输出长度的时候。

就当作给自己记下来好了

Given a string containing digits from 2-9 inclusive, 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. Note that 1 does not map to any letters.

Example:

Input: "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.

解析：

Initial state:

• result = {""}

Stage 1 for number "1":

• result has {""}
• candiate is "abc"
• generate three strings "" + "a", ""+"b", ""+"c" and put into tmp,
  tmp = {"a", "b","c"}
• swap result and tmp (swap does not take memory copy)
• Now result has {"a", "b", "c"}

Stage 2 for number "2":

• result has {"a", "b", "c"}
• candidate is "def"
• generate nine strings and put into tmp,
  "a" + "d", "a"+"e", "a"+"f",
  "b" + "d", "b"+"e", "b"+"f",
  "c" + "d", "c"+"e", "c"+"f"
• so tmp has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }
• swap result and tmp
• Now result has {"ad", "ae", "af", "bd", "be", "bf", "cd", "ce", "cf" }

以次类推

vector<string> letterCombinations(string digits) {
         vector<string> res;
        if(digits.empty())
            return vector<string>();
        
        int len = (int)digits.length();
        static const vector<string> map = {"","","abc", "def", "ghi", "jkl", "mno", "pqrs", "tuv", "wxyz"};
        res.push_back("");
        
        //这应该是一个迭代,好像递归和迭代经常用于不确定大小的东西
        for(int i = 0 ; i < len; i++){
            int num = digits[i] - '0';
            if(num < 0 || num > 9)
                break;
            const string & candidate = map[num];
            if(candidate.empty())
                continue;
            vector<string>tmp;  //每一个循环里面都新建了一个tmp,初始为空，突然觉得循环很像inline函数啊-  -
            for(int j = 0; j < candidate.size(); j++)
                for(int k = 0; k < res.size(); k++)
                    tmp.push_back(res[k] + candidate[j]);
            res.swap(tmp);
            
        }
        
        return res;
    }