#yyds干货盘点# leetcode算法题：解数独

Posted 2022-06-28 灰太狼_cxh

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 1：

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"]]

代码实现：

class Solution 
    private boolean[][] line = new boolean[9][9];
    private boolean[][] column = new boolean[9][9];
    private boolean[][][] block = new boolean[3][3][9];
    private boolean valid = false;
    private List<int[]> spaces = new ArrayList<int[]>();

    public void solveSudoku(char[][] board) 
        for (int i = 0; i < 9; ++i) 
            for (int j = 0; j < 9; ++j) 
                if (board[i][j] == .) 
                    spaces.add(new int[]i, j);
                 else 
                    int digit = board[i][j] - 0 - 1;
                    line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
                
            
        

        dfs(board, 0);
    

    public void dfs(char[][] board, int pos) 
        if (pos == spaces.size()) 
            valid = true;
            return;
        

        int[] space = spaces.get(pos);
        int i = space[0], j = space[1];
        for (int digit = 0; digit < 9 && !valid; ++digit) 
            if (!line[i][digit] && !column[j][digit] && !block[i / 3][j / 3][digit]) 
                line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = true;
                board[i][j] = (char) (digit + 0 + 1);
                dfs(board, pos + 1);
                line[i][digit] = column[j][digit] = block[i / 3][j / 3][digit] = false;