LeetCode - 37. Sudoku Solver

37. Sudoku Solver

Problem's Link

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[b]Mean:[/b]

求解数独.

[b]analyse:[/b]

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.

[b]Time complexity: O(N)[/b]

[b]view code[/b]

/**
* -----------------------------------------------------------------
* 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;
}
/*

*/