LeetCode - 36. Valid Sudoku
2017-03-05 17:27
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36. Valid Sudoku
Problem's Link
----------------------------------------------------------------------------
Mean:
给定一个数独,判断这个数独是否合法.
analyse:
略.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-02-18.28
*/
#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:
bool isValidSudoku(vector<vector<char>>& board)
{
int row=board.size();
int col=board[0].size();
set<int> se;
// -------
for(int i=0;i<row;++i)
{
se.clear();
for(int j=0;j<col;++j)
if(board[i][j]!='.')
{
int k=board[i][j]-'0';
if(se.count(k))
return false;
se.insert(k);
}
}
// -------
for(int i=0;i<col;++i)
{
se.clear();
for(int j=0;j<row;++j)
if(board[j][i]!='.')
{
int k=board[j][i]-'0';
if(se.count(k))
return false;
se.insert(k);
}
}
//*****
for(int i=0;i<row-2;i+=3)
{
for(int j=0;j<col-2;j+=3)
if(!checkBoard(board,i,j))
return false;
}
return true;
}
bool checkBoard(vector<vector<char>>& board,int bi,int bj)
{
set<int> se;
for(int i=bi;i<bi+3;++i)
{
for(int j=bj;j<bj+3;++j)
{
if(board[i][j]!='.')
{
int k=board[i][j]-'0';
if(se.count(k))
return false;
se.insert(k);
}
}
}
return true;
}
};
int main()
{
freopen("H:\\Code_Fantasy\\in.txt","r",stdin);
vector<vector<char>> ve;
string s;
while(cin>>s)
{
vector<char> tempVe;
for(int i=0;s[i];++i)
{
tempVe.push_back(s[i]);
}
ve.push_back(tempVe);
}
Solution solution;
bool ans=solution.isValidSudoku(ve);
puts(ans?"true":"false");
return 0;
}
/*
*/
Problem's Link
----------------------------------------------------------------------------Mean:
给定一个数独,判断这个数独是否合法.
analyse:
略.
Time complexity: O(N)
view code
/**
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-02-18.28
*/
#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:
bool isValidSudoku(vector<vector<char>>& board)
{
int row=board.size();
int col=board[0].size();
set<int> se;
// -------
for(int i=0;i<row;++i)
{
se.clear();
for(int j=0;j<col;++j)
if(board[i][j]!='.')
{
int k=board[i][j]-'0';
if(se.count(k))
return false;
se.insert(k);
}
}
// -------
for(int i=0;i<col;++i)
{
se.clear();
for(int j=0;j<row;++j)
if(board[j][i]!='.')
{
int k=board[j][i]-'0';
if(se.count(k))
return false;
se.insert(k);
}
}
//*****
for(int i=0;i<row-2;i+=3)
{
for(int j=0;j<col-2;j+=3)
if(!checkBoard(board,i,j))
return false;
}
return true;
}
bool checkBoard(vector<vector<char>>& board,int bi,int bj)
{
set<int> se;
for(int i=bi;i<bi+3;++i)
{
for(int j=bj;j<bj+3;++j)
{
if(board[i][j]!='.')
{
int k=board[i][j]-'0';
if(se.count(k))
return false;
se.insert(k);
}
}
}
return true;
}
};
int main()
{
freopen("H:\\Code_Fantasy\\in.txt","r",stdin);
vector<vector<char>> ve;
string s;
while(cin>>s)
{
vector<char> tempVe;
for(int i=0;s[i];++i)
{
tempVe.push_back(s[i]);
}
ve.push_back(tempVe);
}
Solution solution;
bool ans=solution.isValidSudoku(ve);
puts(ans?"true":"false");
return 0;
}
/*
*/
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