《Programming Interviews Exposed》 中的一道递归题：Telephone number to words

题目：

电话键盘上通常一个数字会对应几个字母，如2->"ABC"，3->“DEF”,等等，给定一个电话号码，输出所有可能的字母组合。如"866-2665"的一种字母组合是"TOOCOOL"，要求输出全部。

解法：

与求前排列的递归类似。

#include <iostream>#include <vector>#include <string>using namespace std;class Solution {public:    void TelephoneWords( vector<string> & numberToLetter, string & number )    {        string words = "";        PrintWords(numberToLetter, number, 0, words);        return;    }        void PrintWords( vector<string> & numberToLetter, string & number, size_t start, string & words )    {        if(start == number.length())        {            cout << words << endl;            return;        }        if( number[start]-'0' >= 0 )        {            for(size_t i = 0; i < numberToLetter[number[start]-'0'].length(); i++)            {                words.insert(words.end(), numberToLetter[number[start]-'0'][i]);                PrintWords(numberToLetter, number, start+1, words);                words.erase(words.length()-1, 1);            }        }        else{            words.insert(words.end(), '-');            PrintWords(numberToLetter, number, start+1, words);            words.erase(words.length()-1, 1);        }    }};int main(int argc, const char * argv[]){    // insert code here...    vector<string> numberToLetter;    numberToLetter.push_back("0");    numberToLetter.push_back("1");    numberToLetter.push_back("ABC");    numberToLetter.push_back("DEF");    numberToLetter.push_back("GHI");    numberToLetter.push_back("JKL");    numberToLetter.push_back("MNO");    numberToLetter.push_back("PRS");    numberToLetter.push_back("TUV");    numberToLetter.push_back("WXY");        string number = "497-1927";        Solution s;    s.TelephoneWords(numberToLetter, number);    return 0;}