37. Sudoku Solver
题目
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.
解析
- 解题思路：深度遍历的过程，从开始一直扫描，直到遇到空字符，便可开始填数字，该位置的数字从1~9依次遍历，不行的时候就回退
// 37. Sudoku Solver
class Solution_37 {
public:
bool isValid(vector<vector<char>>& board,int i,int j) //只需要判断当前行，列，方格是否合法，节省时间
{
for (int row = 0; row < 9;row++)
{
if (row!=i&&board[i][j]==board[row][j])
{
return false;
}
}
for (int col = 0; col < 9;col++)
{
if (col!=j&&board[i][j]==board[i][col])
{
return false;
}
}
for (int row = i / 3 * 3; row < i / 3 * 3 + 3;row++)
{
for (int col = j / 3 * 3; col < j / 3 * 3 + 3;col++)
{
if ((row!=i||col!=j)&&board[i][j]==board[row][col]) //小方格也需要行列判断
{
return false;
}
}
}
return true;
}
bool dfs(vector<vector<char> > &board,int i,int j)
{
if (i==9)
{
return true;
}
if (j==9)
{
return dfs(board, i + 1, 0);
}
if (board[i][j]=='.')
{
for (char k = '1'; k <='9';k++)
{
board[i][j] = k;
if (isValid(board,i,j))
{
if (dfs(board, i, j + 1))
{
return true;
}
}
board[i][j] = '.'; //达到回溯的目的
}
}
else
{
return dfs(board, i, j + 1);
}
return false;
}
void solveSudoku(vector<vector<char>>& board) {
if (board.size()<9||board[0].size()<9)
{
return;
}
dfs(board, 0, 0);
return;
}
};
题目来源
