LeetCode - 17. Letter Combinations of a Phone Number

 17. Letter Combinations of a Phone Number

Problem's Link

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Mean: 

给你一个数字串，输出其在手机九宫格键盘上的所有可能组合.

analyse:

使用进位思想来枚举所有组合即可.

Time complexity: O(N)

 

view code

/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
*       Author: crazyacking
*       Date  : 2016-02-17-09.57
*/
#include <queue>
#include <cstdio>
#include <set>
#include <string>
#include <stack>
#include <cmath>
#include <climits>
#include <map>
#include <cstdlib>
#include <iostream>
#include <vector>
#include <algorithm>
#include <cstring>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);

class Solution
{
public :
    map < char , string > mp;
    vector < string > letterCombinations( string digits)
    {
        mp [ '0' ] = string( " ");
        mp [ '1' ] = string( "");
        mp [ '2' ] = string( "abc");
        mp [ '3' ] = string( "def");
        mp [ '4' ] = string( "ghi");
        mp [ '5' ] = string( "jkl");
        mp [ '6' ] = string( "mno");
        mp [ '7' ] = string( "pqrs");
        mp [ '8' ] = string( "tuv");
        mp [ '9' ] = string( "wxyz");

        vector < int > cntBase;
        vector < string > res;
        int len = digits . length();
        if( len <= 0) return res;
        for( int i = 0; i < len; ++ i)
            cntBase . push_back( 0);
       
        string tmp;
        while( cntBase [ 0 ] < mp [ digits . at( 0 )]. length())
        {
            tmp . clear();
            for( int i = 0; i < len; ++ i)
            {
                tmp . push_back( mp [ digits . at( i )][ cntBase [ i ]]);
            }
            res . push_back( tmp);
            cntBase [ len - 1 ] ++;
            // check and carry
            int carry = 0;
            for( int i = len - 1; i > 0; -- i)
            {
                cntBase [ i ] += carry;
                carry = cntBase [ i ] / ( mp [ digits . at( i )]. length());
                cntBase [ i ] %=( mp [ digits . at( i )]. length());
            }
            cntBase [ 0 ] += carry;
        }
        return res;
    }
};
int main()
{
    Solution solution;
    string s;
    while( cin >>s)
    {
        auto ve = solution . letterCombinations(s);
        for( auto p: ve)
            cout <<p << endl;
    }
    return 0;
}
/*

*/