37. Sudoku Solver
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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; } /* */