LC-37. 解数独

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37. 解数独

一、题目描述： 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 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"]]
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、思路分析：

回溯

典型的回溯问题， 对每一个需要填入数字的位置尝试填入1-9，如果发现填入某个数会导致数独解不下去，则进行回溯。

• 终止条件： 数独填写完毕
• 可以做出的选择： 尝试填入1-9
• 已经做出的选择：更新数独数组

三、AC 代码：

class Solution {
    boolean[][] row, col;
    boolean[][][] cell;

    public void solveSudoku(char[][] board) {
        this.row = new boolean[9][10];
        this.col = new boolean[9][10];
        this.cell = new boolean[3][3][10];
        for (int i = 0; i < board.length; i++) {
            for (int j = 0; j < board[0].length; j++) {
                if (board[i][j] == '.') continue;
                int v = board[i][j] - '0';
                 row[i][v] = col[j][v] = cell[i / 3][j / 3][v] = true;
            }
        }
        dfs(board,0,0);
    }

    boolean dfs(char[][] board, int i, int j) {
        if (i == board.length) return true;
        if (j == board[0].length) return dfs(board, i + 1, 0);
        if (board[i][j] != '.') return dfs(board, i, j + 1);
        for (int t = 1; t <= 9; t++) {
            if (row[i][t] || col[j][t] || cell[i / 3][j / 3][t]) continue;
            board[i][j] = (char) ('0' + t);
            row[i][t] = col[j][t] = cell[i / 3][j / 3][t] = true;
            boolean ans = dfs(board, i, j + 1);
            if (ans) return true;
            row[i][t] = col[j][t] = cell[i / 3][j / 3][t] = false;
            board[i][j] = '.';
        }

        return false;
    }
}