LeetCode——037

/*

37. Sudoku Solver My Submissions QuestionEditorial Solution

Total Accepted: 48543 Total Submissions: 195911 Difficulty: Hard

Write a program to solve a Sudoku puzzle by filling the empty cells.

Empty cells are indicated by the character ‘.’.

You may assume that there will be only one unique solution.



*/

/*

解题思路：

采用递归的方法，递归函数在这里有两重作用，返回值代表从当前状态到继续下去的话是否可行，而参数中引用记录当前状态即携带最终的结果。对于每个需要填数字的格子带入1到9，每代入一个数字都判定其是否合法，如果合法就继续下一次递归，结束时把数字设回’.’，判断新加入的数字是否合法时，只需要判定当前数字是否合法，不需要判定这个数组是否为数独数组，因为之前加进的数字都是合法的，这样可以使程序更加高效一些，具体实现如代码所示：

*/

class Solution {
public:
void solveSudoku(vector<vector<char>>& board) {

if(board.empty()||board.size()!=9 || board[0].size()!=9)return ;
solveSudoku(board,0,0);

}

bool solveSudoku(vector<vector<char>>& board,int i,int j){

if(i==9)return true;
if(j>=9)return solveSudoku(board,i+1,0);
if(board[i][j]=='.'){
for(int k=1;k<=9;k++){
board[i][j]=(char)(k+'0');
if(isvalid(board,i,j)&& solveSudoku(board,i,j+1)){

return true;
}
board[i][j]='.';

}
}else return solveSudoku(board,i,j+1);
return false;
}

bool isvalid(vector<vector<char>> &board,int i,int j){

//判断行
for(int col=0;col<9;++col){

if(col==j)continue;
if(board[i][col]==board[i][j])return false;
}
//判断列
for(int row=0;row<9;++row){

if(row==i)continue;
if(board[row][j]==board[i][j])return false;
}

//判断小方阵
for(int r=i/3*3;r<i/3*3+3;++r){
for(int c=j/3*3;c<j/3*3+3;++c){

if((r!=i||c!=j)&&board[i][j]==board[r][c])return false;
}
}

return true;
}

};