LeetCode - 17. Letter Combinations of a Phone Number
17. Letter Combinations of a Phone Number#
Problem's Link
#
----------------------------------------------------------------------------
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)
/**
* -----------------------------------------------------------------
* 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;
}
/*
*/
* -----------------------------------------------------------------
* 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;
}
/*
*/
作者:北岛知寒
出处:https://www.cnblogs.com/crazyacking/p/5194906.html
版权:本作品采用「署名-非商业性使用-相同方式共享 4.0 国际」许可协议进行许可。
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