17.leetcode Letter Combinations of a Phone Number(meidum)[递归回溯]

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"].

这道题与常见的递归回溯没有区别，但是需要注意的是每个数字对应的字符与ascii之间的转换关系，当字符为7和9时字符个数变成了4，因此后面的计算字符的公式需要 重新定义。

class Solution {public:    void getLetter(int start,int n,string &temp,string &digit,vector<string> &result)    {        if(start == n)        {            if(temp != "")                result.push_back(temp);            return ;        }        int k = 3;        for(int i = 0;i<k;i++)        {            char c = digit[start];            if(c<='1'||c>'9') return ;            if(c == '7' ||c == '9') k = 4;            char d;            if(c == '8' || c == '9')                d = c+(c-'2')*2+47+i+1;            else                d = c+(c-'2')*2+47+i;            string bef = temp;            temp += d;            getLetter(start+1,n,temp,digit,result);            temp = bef;        }        return;    }    vector<string> letterCombinations(string digits) {        int n = digits.size();        vector<string> result;        if(n <=0) return result;        string temp = "";        getLetter(0,n,temp,digits,result);        return result;    }};