LeetCode No37. 解数独

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"]]
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

思路

其实，如果知道上一题的哈希解法，这道题目就很简单了，回溯直接解决。

AC代码

点击查看代码

class Solution {
    boolean[][] rowOk = new boolean[10][10];
    boolean[][] colOk = new boolean[10][10];
    boolean[][][] littleSudo = new boolean[3][3][10];
    boolean end = false;
    List<Integer> blankPos = new ArrayList<Integer>();

    public void DFS(char[][] board, int cur) {
        if( cur==blankPos.size() ) {
            end = true;
            return ;
        }
        int pos = blankPos.get(cur);
        int i = pos/10;
        int j = pos%10;
        for(int num=1; num<10 && !end; num++) {
            if( !rowOk[i][num] && !colOk[j][num] && !littleSudo[i/3][j/3][num] ) {
                rowOk[i][num] = true;
                colOk[j][num] = true;
                littleSudo[i/3][j/3][num] = true;
                board[i][j] = (char) (num+'0');
                DFS(board, cur+1);
                rowOk[i][num] = false;
                colOk[j][num] = false;
                littleSudo[i/3][j/3][num] = false;
            }
        }
    }

    public void solveSudoku(char[][] board) {
        for(int i=0; i<9; i++) {
            for(int j=0; j<9; j++) {
                if( board[i][j] == '.' ) {
                    blankPos.add(i*10+j);
                } else {
                    int num = (int) (board[i][j]-'0');
                    rowOk[i][num] = true;
                    colOk[j][num] = true;
                    littleSudo[i/3][j/3][num] = true;
                }
            }
        }
        DFS(board, 0);
    }
}