leetcode 17. Letter Combinations of a Phone Number

题目描述：
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.

讲道理一开始还没看懂，就是按照手机的数字键盘上面对应的字母，将数字串转换为可能的字母串。看了下面这个表就明白了。

1       2(abc)  3(def)
4(ghi)  5(jkl)  6(mno)
7(pqrs) 8(tuv)  9(wxyz)
*       0       #

应该不难想到可以用递归的方法一层一层循环的遍历。时间复杂度是O(n*m)，以下是解答的方法：

class Solution {
private:
    vector<string> result;
public:
    vector<string> letterCombinations(string digits) {
        if (digits == "") {
            return result;
        }
        recursion(digits, 0, "");
        return result;
    }
    void recursion(string digits, int i,string now) { 

        string digitToletter[8] = { "abc", "def","ghi","jkl","mno","pqrs","tuv","wxyz"};

        int letterNum = 3;
        if (digits[i] == '7' || digits[i] == '9') {
            letterNum = 4;
        }
            for (int k = 0; k < letterNum; k++) {
                if (i == 0) {
                    now = "";
                }
                now = now + digitToletter[digits[i]-'2'][k];
                if (i == digits.length() - 1) {
                    result.push_back(now);
                }
                else {
                    recursion(digits, i + 1, now);
                }
                //这里的两次字符串长度减一是十分关键的，
                //这里是遍历完一个数字对应的一个字母后，字符串长度减一以便继续遍历
                now = now.substr(0, now.length() - 1);
            }
            //这里是遍历完一个数字对应的所有字母后，回到上一层的数字进行遍历
            now = now.substr(0, now.length() - 1);
        }

};