思路:
DFS。时间复杂度O(9^4),空间复杂度O(1)。
下棋添子,下完一步下另一步,不行就回溯。
class Solution {
private:
bool isValid(vector<vector<char>> &board, int i, int j) {
//check col
for(int row = 0; row < 9; ++row) {
if(row != i && board[row][j] == board[i][j]) {
return false;
}
}
//check row
for(int col = 0; col < 9; ++col) {
if(col != j && board[i][col] == board[i][j]) {
return false;
}
}
//check cube
for(int row = i - i%3; row < i + (3 - i%3); ++row) {
for(int col = j - j%3; col < j + (3 - j%3); ++col) {
if(row == i && col == j) continue;
if(board[row][col] == board[i][j]) {
return false;
}
}
}
return true;
}
bool dfs(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] = '0' + k;
if(isValid(board, i, j) && dfs(board)) {
return true;
}
board[i][j] = '.';
}
return false;
}
}
}
return true;
}
public:
void solveSudoku(vector<vector<char>>& board) {
dfs(board);
}
};