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37. Sudoku Solver Problem's Link
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Mean:
求解数独.
analyse:
只是9宫格的数独,而且测试数据都不难,所以可以直接使用递归求解,类似于N-Queue问题.
但如果宫格数较多,则需要使用Dancing-Link精确覆盖算法来求解.
Time complexity: O(N)
view code
* -----------------------------------------------------------------
* Copyright (c) 2016 crazyacking.All rights reserved.
* -----------------------------------------------------------------
* Author: crazyacking
* Date : 2016-03-02-18.53
*/
#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>
#include <bits/stdc++.h>
#include <windows.h>
using namespace std;
typedef long long( LL);
typedef unsigned long long( ULL);
const double eps( 1e-8);
class Solution
{
public :
void solveSudoku( vector < vector < char >>& board)
{
recursiveSolve( board);
}
bool recursiveSolve( vector < vector < char >>& board)
{
for( int i = 0; i < 9; ++ i)
{
for( int j = 0; j < 9; ++ j)
{
if( board [ i ][ j ] == '.')
{
for( int k = 1; k <= 9; ++ k)
{
board [ i ][ j ] = static_cast < char >( k + '0');
if( isValid( board , i , j) && recursiveSolve( board))
return true;
board [ i ][ j ] = '.';
}
return false;
}
}
}
return true;
}
bool isValid( const vector < vector < char >>& board , const int r1 , const int c1) const
{
for( int i = 0; i < 9; ++ i)
{
if( i != r1 && board [ i ][ c1 ] == board [ r1 ][ c1 ])
return false;
if( i != c1 && board [ r1 ][ i ] == board [ r1 ][ c1 ])
return false;
}
int rowBegin = r1 / 3 * 3;
int colBegin = c1 / 3 * 3;
for( int i = rowBegin; i < rowBegin + 3; ++ i)
{
for( int j = colBegin; j < colBegin + 3; ++ j)
{
if( i != r1 && j != c1 && board [ i ][ j ] == board [ r1 ][ c1 ])
return false;
}
}
return true;
}
};
int main()
{
freopen( "H: \\ Code_Fantasy \\ in.txt" , "r" , stdin);
Solution solution;
vector < vector < char >> ve;
string s;
while( cin >>s)
{
vector < char > tempVe;
for( int i = 0; i <s . length(); ++ i)
tempVe . push_back(s [ i ]);
ve . push_back( tempVe);
}
solution . solveSudoku( ve);
return 0;
}
/*
*/
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