一、题目
编写一个程序，通过填充空格来解决数独问题。
数独的解法需 遵循如下规则：
数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。（请参考示例图）
数独部分空格内已填入了数字，空白格用 '.' 表示。
示例：
输入：board = [["5","3",".",".","7",".",".",".","."],["6",".",".","1","9","5",".",".","."],[".","9","8",".",".",".",".","6","."],["8",".",".",".","6",".",".",".","3"],["4",".",".","8",".","3",".",".","1"],["7",".",".",".","2",".",".",".","6"],[".","6",".",".",".",".","2","8","."],[".",".",".","4","1","9",".",".","5"],[".",".",".",".","8",".",".","7","9"]]
输出：[["5","3","4","6","7","8","9","1","2"],["6","7","2","1","9","5","3","4","8"],["1","9","8","3","4","2","5","6","7"],["8","5","9","7","6","1","4","2","3"],["4","2","6","8","5","3","7","9","1"],["7","1","3","9","2","4","8","5","6"],["9","6","1","5","3","7","2","8","4"],["2","8","7","4","1","9","6","3","5"],["3","4","5","2","8","6","1","7","9"]]
解释：输入的数独如上图所示，唯一有效的解决方案如下所示：
提示：
board.length == 9
board[i].length == 9
board[i][j] 是一位数字或者 '.'
题目数据 保证 输入数独仅有一个解
二、思路
- 二进制优化， 一共有9个数字，那么用9位二进制代表每个数，若该位置为1代表这个数可以在该位置填入。
- 如何进行剪枝呢，我们可以从搜索分支少的点开始进行搜索，开始的分支少，那么搜索的次数便会减少。每个位置可以搜索的点便是grid ， row, col三个数组的并集，二进制上的1代表可以填，那么3个数组都满足就是最后能填入的数。
- lowbit获取可以填入1的位置，遍历所有的1即是能填入的数。
class Solution {
public:
int cnt = 81;
void solveSudoku(vector<vector<char>>& board) {
vector<vector<int>> grid(9, vector<int>(10));
vector<vector<int>> row(9, vector<int>(10));
vector<vector<int>> col(9, vector<int>(10));
init(board, grid, row, col);
dfs(board, grid, row, col, 0, 0);
}
void init(vector<vector<char>>& board, vector<vector<int>>& grid, vector<vector<int>>& row, vector<vector<int>>& col) {
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
int idx = i / 3 * 3 + j / 3;
int t = board[i][j] - '0';
grid[idx][t] = 1;
row[i][t] = 1;
col[j][t] = 1;
cnt--;
}
}
}
}
bool dfs(vector<vector<char>>& board, vector<vector<int>>& grid, vector<vector<int>>& row, vector<vector<int>>& col, int i, int j) {
if (cnt == 0) return true;
//每次从下一个位置开始搜索
while (board[i][j] != '.') {
if (++j >= 9) {
i++;
j = 0;
}
if (i >= 9) return true;
}
if (board[i][j] == '.') {
for (int k = 1; k <= 9; k++) {
int idx = i / 3 * 3 + j / 3;
if (!grid[idx][k] && !row[i][k] && !col[j][k]) {
board[i][j] = char(k + '0');
grid[idx][k] = 1;
row[i][k] = 1;
col[j][k] = 1;
cnt--;
//printf("%d-%d-%c\n", i + 1, j + 1, board[i][j]);
if (dfs(board, grid, row, col, i, j)) return true;
cnt++;
board[i][j] = '.';
grid[idx][k] = 0;
row[i][k] = 0;
col[j][k] = 0;
}
}
}
return false;
}
};
- 二进制优化
class Solution {
public:
unordered_map<int, int> mp;
void solveSudoku(vector<vector<char>>& board) {
vector<int> grid(9);
vector<int> row(9);
vector<int> col(9);
vector<int> help((1 << 9));
int cnt = 81;
vector<vector<int>> g;
for (int i = 0; i < 9; i++) {
mp[1 << i] = i;
grid[i] = (1 << 9) - 1;
row[i] = (1 << 9) - 1;
col[i] = (1 << 9) - 1;
}
//统计每个二进制有多少个1
for (int i = 0; i < (1 << 9); i++) {
int cnt = 0;
for (int j = i; j; j -= lowbit(j)) cnt++;
help[i] = cnt;
}
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] != '.') {
int t = board[i][j] - '1';
int idx = i / 3 * 3 + j / 3;
grid[idx] -= 1 << t;
row[i] -= 1 << t;
col[j] -= 1 << t;
cnt--;
}
}
}
dfs(board, grid, row, col, help, cnt);
}
int lowbit(int x) {
return x & (-x);
}
int get(int i, int j, vector<int>& grid, vector<int>& row, vector<int>& col) {
return grid[i / 3 * 3 + j / 3] & row[i] & col[j];
}
bool dfs(vector<vector<char>>& board, vector<int>& grid, vector<int>& row, vector<int>& col, vector<int>& help, int ct) {
//找出搜索范围最小的位置进行填数 即1个数最少的点
if (ct == 0) return true;
int x = 0, y = 0, minv = 20;
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
if (board[i][j] == '.') {
int cnt = help[get(i, j, grid, row, col)];
if (cnt < minv) {
x = i, y =j;
minv = cnt;
}
}
}
}
//从(x,y)这个位置开始进行搜索
for (int j = get(x, y, grid, row, col); j; j -= lowbit(j)) {
grid[x / 3 * 3 + y / 3] -= lowbit(j);
row[x] -= lowbit(j);
col[y] -= lowbit(j);
board[x][y] = char(mp[lowbit(j)] + '1');
if (dfs(board, grid, row, col, help, ct - 1)) return true;
board[x][y] = '.';
grid[x / 3 * 3 + y / 3] += lowbit(j);
row[x] += lowbit(j);
col[y] += lowbit(j);
}
return false;
}
};
