英文原文
Write a program to solve a Sudoku puzzle by filling the empty cells.
A sudoku solution must satisfy all of the following rules:
- Each of the digits
1-9
must occur exactly once in each row. - Each of the digits
1-9
must occur exactly once in each column. - Each of the digits
1-9
must occur exactly once in each of the 93x3
sub-boxes of the grid.
The '.'
character indicates empty cells.
Example 1:
Input: 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"]] Output: [["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"]] Explanation: The input board is shown above and the only valid solution is shown below:
Constraints:
board.length == 9
board[i].length == 9
board[i][j]
is a digit or'.'
.- It is guaranteed that the input board has only one solution.
中文题目
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
通过代码
高赞题解
思路
状态压缩
使用
bitset<9>
来压缩存储每一行、每一列、每一个3x3
宫格中1-9
是否出现这样每一个格子就可以计算出所有不能填的数字,然后得到所有能填的数字
getPossibleStatus()
填入数字和回溯时,只需要更新存储信息
每个格子在使用时,会根据存储信息重新计算能填的数字
回溯
每次都使用
getNext()
选择能填的数字最少的格子开始填,这样填错的概率最小,回溯次数也会变少使用
fillNum()
在填入和回溯时负责更新存储信息一旦全部填写成功,一路返回 true ,结束递归
图解
答题
class Solution {
public:
bitset<9> getPossibleStatus(int x, int y)
{
return ~(rows[x] | cols[y] | cells[x / 3][y / 3]);
}
vector<int> getNext(vector<vector<char>>& board)
{
vector<int> ret;
int minCnt = 10;
for (int i = 0; i < board.size(); i++)
{
for (int j = 0; j < board[i].size(); j++)
{
if (board[i][j] != '.') continue;
auto cur = getPossibleStatus(i, j);
if (cur.count() >= minCnt) continue;
ret = { i, j };
minCnt = cur.count();
}
}
return ret;
}
void fillNum(int x, int y, int n, bool fillFlag)
{
rows[x][n] = (fillFlag) ? 1 : 0;
cols[y][n] = (fillFlag) ? 1 : 0;
cells[x/3][y/3][n] = (fillFlag) ? 1: 0;
}
bool dfs(vector<vector<char>>& board, int cnt)
{
if (cnt == 0) return true;
auto next = getNext(board);
auto bits = getPossibleStatus(next[0], next[1]);
for (int n = 0; n < bits.size(); n++)
{
if (!bits.test(n)) continue;
fillNum(next[0], next[1], n, true);
board[next[0]][next[1]] = n + '1';
if (dfs(board, cnt - 1)) return true;
board[next[0]][next[1]] = '.';
fillNum(next[0], next[1], n, false);
}
return false;
}
void solveSudoku(vector<vector<char>>& board)
{
rows = vector<bitset<9>>(9, bitset<9>());
cols = vector<bitset<9>>(9, bitset<9>());
cells = vector<vector<bitset<9>>>(3, vector<bitset<9>>(3, bitset<9>()));
int cnt = 0;
for (int i = 0; i < board.size(); i++)
{
for (int j = 0; j < board[i].size(); j++)
{
cnt += (board[i][j] == '.');
if (board[i][j] == '.') continue;
int n = board[i][j] - '1';
rows[i] |= (1 << n);
cols[j] |= (1 << n);
cells[i / 3][j / 3] |= (1 << n);
}
}
dfs(board, cnt);
}
private:
vector<bitset<9>> rows;
vector<bitset<9>> cols;
vector<vector<bitset<9>>> cells;
};
感谢 @cocowy 提供的 bitset 风格的 fillNum()
,可读性得到质的提升,已更新到上面
下面是我原来位运算风格的 fillNum()
void fillNum(int x, int y, int n, bool fillFlag)
{
bitset<9> pick(1 << n);
rows[x] = (fillFlag) ? (rows[x] | pick) : (rows[x] ^ pick);
cols[y] = (fillFlag) ? (cols[y] | pick) : (cols[y] ^ pick);
cells[x / 3][y / 3] = (fillFlag) ? (cells[x / 3][y / 3] | pick) : (cells[x / 3][y / 3] ^ pick);
}
执行时间
对格子填入顺序的选择,使得搜索效率大大提升(加上运气好),击败 100%
得益于状态压缩,内存的使用也击败 100%
(别太较真儿,看个乐就好)
致谢
感谢您的观看,希望对您有帮助,欢迎热烈的交流!
如果感觉还不错就点个赞吧~
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