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37-解数独(Sudoku Solver)
发表于:2021-12-03 | 分类: 困难
字数统计: 1.7k | 阅读时长: 9分钟 | 阅读量:

原文链接: https://leetcode-cn.com/problems/sudoku-solver

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:

  1. Each of the digits 1-9 must occur exactly once in each row.
  2. Each of the digits 1-9 must occur exactly once in each column.
  3. Each of the digits 1-9 must occur exactly once in each of the 9 3x3 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

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 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

  1. 状态压缩

    1. 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

    2. 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()

    3. 填入数字和回溯时,只需要更新存储信息

    4. 每个格子在使用时,会根据存储信息重新计算能填的数字

  1. 回溯

    1. 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

    2. 使用 fillNum() 在填入和回溯时负责更新存储信息

    3. 一旦全部填写成功,一路返回 true ,结束递归

图解

图片.png

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

图片.png

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~

统计信息

通过次数 提交次数 AC比率
109590 163431 67.1%

提交历史

提交时间 提交结果 执行时间 内存消耗 语言

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