题目链接：

力扣

题目详情：

37. 解数独t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

思路：

我们使用三个boolean数组，多开一个位置让下标与数字映射

rowCheck = new boolean[9][10];

colCheck = new boolean[9][10];

check = new boolean[3][3][10];

• boolean[][] rowCheck,用来表示某一行是否能填写某个数字：比如rowCheck[3][4]就表示下标为3这一行是否存在4这个数字，如果为false，就表示不存在
• boolean[][] colCheck;同上，表示某一列是否能填写某个数字
• boolean[][][] check;用来表示某一个小方块是否能填写某个数字，像下面这样把数度分为九个小方块，我们可以使用横坐标以及纵坐标/3的值来表示具体是哪个方块，比如下标为【3】【3】就可以使用check[3/1][3/1]来表示，那么check[3/1][3/1][4]就用来表示第一个小方块中是否存在4这个数字

确定好了Boolean数组的表示含义，我们就可以来写代码了，在写代码之前，我们需要先遍历数独中写好的数据，修改我们的Boolean数组，代码如下：

for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {if (board[i][j] != '.') {int num = board[i][j] - '0';rowCheck[i][num] = colCheck[j][num] = check[i/3][j/3][num] = true;}}}

这样在调用我们写的dfs（char[][] board）方法，

完整代码（有详细注释）：

class Solution {boolean[][] rowCheck, colCheck;boolean[][][] check;public void solveSudoku(char[][] board) {rowCheck = new boolean[9][10];colCheck = new boolean[9][10];check = new boolean[3][3][10];// 遍历数独初始化Boolean数组for (int i = 0; i < 9; i++) {for (int j = 0; j < 9; j++) {if (board[i][j] != '.') {int num = board[i][j] - '0';rowCheck[i][num] = colCheck[j][num] = check[i/3][j/3][num] = true;}}}dfs(board);}public boolean dfs(char[][] board) {for (int row = 0; row < 9; row++) {for (int col = 0; col < 9; col++) {// 如果等于'.'说明需要填写数字if (board[row][col] == '.') {// 遍历数字1-9,选取合适的for (int num = 1; num < 10; num++) {// 如果行列以及所处小方块都不存在当前的数字，说明可以放if (!rowCheck[row][num] && !colCheck[col][num] && !check[row/3][col/3][num]) {// 修改当前的数字board[row][col] = (char)(num+'0');// 修改当前行列以及小方块的Boolean值rowCheck[row][num] = colCheck[col][num] = check[row/3][col/3][num] = true;// 如果接下来返回的为true,说明当前选择是正确的if (dfs(board)) {return true; } else {// 否则说明选择错误，就需要回溯到指定的位置，重新进行选择board[row][col] = '.';// 修改Boolean值回到初始状态rowCheck[row][num] = colCheck[col][num] = check[row/3][col/3][num] = false;}}}// 走到这里说明数字1-9一个也不能填，直接返回falsereturn false;}}}//两层循环都走完说明数独全部填完，返回truereturn true;}
}