Careercup

2014-05-02 10:07

题目链接

原题：

Mapping 
'1' = 'A','B','C' 
'2' = 'D','E','F' 
... 
'9' = 

input: 112 
output :ouput = [AAD, BBD, CCD, AAE, AAF, BBE, BBF, CCE, CCF]

题目：电话的数字按键和字母有个映射关系，给定一串数字，请给出所有可能的字符映射方式。

解法：此人也不给个描述，搞的下面一堆人来猜题意。这个题目的意思是说，对于给定的数字串，有多少种不同的映射方式。像“112” -> “ABD”这样是不允许的，因为“1”只能表示一种字母，不能同时表示“A”和“B”。

代码：

// http://www.careercup.com/question?id=5765850736885760
#include <iostream>
#include <string>
#include <vector>
using namespace std;

void DFS(int idx, vector<pair<int, char> > &dict, vector<string> &result, const string &s)
{
    int i;
    
    if (idx == (int)dict.size()) {
        static char m[10];
        
        for (i = 0; i < (int)dict.size(); ++i) {
            m[dict[i].first] = dict[i].second;
        }
        
        static string res;
        
        res.clear();
        for (i = 0; i < (int)s.length(); ++i) {
            res.push_back(m[s[i] - '1']);
        }
        
        result.push_back(res);
        return;
    }
    
    switch(dict[idx].first) {
    case 0:
    case 1:
    case 2:
    case 3:
    case 4:
    case 5:
    case 6:
    case 7:
        for (i = 0; i < 3; ++i) {
            dict[idx].second = (dict[idx].first * 3 + i) + 'A';
            DFS(idx + 1, dict, result, s);
        }
        break;
    case 8:
        for (i = 0; i < 2; ++i) {
            dict[idx].second = (dict[idx].first * 3 + i) + 'A';
            DFS(idx + 1, dict, result, s);
        }
        break;
    }
}

int main()
{
    int i;
    string s;
    vector<pair<int, char> > dict;
    vector<string> result;
    int a[10];
    
    while (cin >> s) {
        for (i = 0; i < 10; ++i) {
            a[i] = 0;
        }
        for (i = 0; i < (int)s.length(); ++i) {
            a[s[i] - '1'] = 1;
        }
        
        for (i = 0; i < 10; ++i) {
            if (a[i]) {
                dict.push_back(make_pair(i, ''));
            }
        }
        
        DFS(0, dict, result, s);
        cout << "{" << endl;
        for (i = 0; i < (int)result.size(); ++i) {
            cout << "    " << result[i] << endl;
        }
        cout << "}" << endl;
        
        dict.clear();
        result.clear();
    }
    
    return 0;
}