37. 解数独

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

思路

使用回溯模板：

void backtracking(参数) {
    if (终止条件) {
        存放结果;
        return;
    }
    for (选择：本层集合中元素（树中节点孩子的数量就是集合的大小）) {
        处理节点;
        backtracking(路径，选择列表); // 递归
        回溯，撤销处理结果
    }
}

本题与 51. N 皇后类似，只是N皇后每行只放置一个皇后，而数独每行的每个空格都要填好，所以需要进行二维递归判断。

一个for循环遍历棋盘的行，一个for循环遍历棋盘的列，一行一列确定下来之后，递归遍历这个位置放9个数字的可能性！

注意这里return false的地方，这里放return false 是有讲究的。

因为如果一行一列确定下来了，这里尝试了9个数都不行，说明这个棋盘找不到解决数独问题的解！

那么会直接返回， 这也就是为什么没有终止条件也不会永远填不满棋盘而无限递归下去！

解法

class Solution {

    public void solveSudoku(char[][] board) {
        backtracking(board);
    }

    public boolean backtracking(char[][] board) {
        // 遍历行
        for (int row = 0; row < board.length; row++) {
            // 遍历列
            for (int col = 0; col < board.length; col++) {
                if (board[row][col] != '.') {
                    continue;
                }
                // (row, col) 这个位置放 i 是否合适
                for (char i = '1'; i <= '9'; i++) {
                    if (check(board, row, col, i)) {
                        // 放置 i
                        board[row][col] = i;
                        if (backtracking(board)) {
                            // 如果找到合适一组立刻返回
                            return true;
                        }
                        // 回溯，撤销
                        board[row][col] = '.';
                    }
                }
                // 9个数都试完了，都不行，那么就返回false
                return false;
            }
        }
        // 遍历完没有返回false，说明找到了合适棋盘位置了
        return true;
    }

    public boolean check(char[][] board, int row, int col, char target) {
        // 同行不重复
        for (int i = 0; i < board.length; i++) {
            if (board[i][col] == target) {
                return false;
            }
        }
        // 同列不重复
        for (int i = 0; i < board.length; i++) {
            if (board[row][i] == target) {
                return false;
            }
        }
        // 3*3方格不重复
        int startRow = row / 3 * 3;
        int startCol = col / 3 * 3;
        for (int i = startRow; i < startRow + 3; i++) {
            for (int j = startCol; j < startCol + 3; j++) {
                if (board[i][j] == target) {
                    return false;
                }
            }
        }
        return true;
    }
}