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);        }};