37. 解数独 - 力扣（LeetCode）

一、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：

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、代码

class Solution {public void solveSudoku(char[][] board) {// row[x][y]：y这个数在第x行是否出现过boolean[][] row = new boolean[9][10];// col[x][y]：y这个数在第x列是否出现过boolean[][] col = new boolean[9][10];// bucket[x][y]：y这个数在第x号宫格内是否出现过boolean[][] bucket = new boolean[9][10];// 初始化上面的三个数组initMaps(board, row, col, bucket);// 开始暴力递归尝试所有答案，只要找到了一种答案就返回process(board, 0, 0, row, col, bucket);}// 初始化三个辅助数组public void initMaps(char[][] board, boolean[][] row, boolean[][] col, boolean[][] bucket) {// 开始遍历数独表，构造辅助数组for (int i = 0; i < board.length; i++) {for (int j = 0; j < board[0].length; j++) {if (board[i][j] != '.') {int num = board[i][j] - '0';int bid = (i / 3) * 3 + (j / 3);// 标记该数出现在的位置row[i][num] = true;col[j][num] = true;bucket[bid][num] = true;}}}}// DFS// 当前来到(i,j)这个位置，// 如果已经有数字，跳到下一个位置上// 如果没有数字，尝试1~9，不能和row、col、bucket冲突public boolean process(char[][] board, int i, int j, boolean[][] row, boolean[][] col, boolean[][] bucket) {// 如果行数已经越界，说明已经尝试完了所有位置，返回trueif (i > 8) {return true;}// 当离开(i，j)，应该去哪？(nexti, nextj)// 要通过j是否走到结尾了来决定是不是i要向下走一行，j是否要重新回到0int nexti = j < 8 ? i : i + 1;int nextj = j < 8 ? j + 1 : 0;// 如果当前位置不是空，就继续尝试下一个位置if (board[i][j] != '.') {return process(board, nexti, nextj, row, col, bucket);// 当前位置是空，我们来尝试1~9} else {// 计算当前数字所在的3*3宫格的编号int bid = (i / 3) * 3 + (j / 3);// 在该位置尝试填写1~9，看能不能符合要求for (int num = 1; num <= 9; num++) {// 如果尝试的数字在该行、列、宫格内从未出现，说明符合要求if (!row[i][num] && !col[j][num] && !bucket[bid][num]) {// 可以尝试numrow[i][num] = true;col[j][num] = true;bucket[bid][num] = true;// 将num填到board中board[i][j] = (char)(num + '0');// 继续递归尝试下一个位置，如果后面都是成立的，就说明找到了一种解法，直接返回true。不再执行后续的尝试，因为题目保证了输入数独仅有一个解if (process(board, nexti, nextj, row, col, bucket)) {return true;}// 恢复现场row[i][num] = false;col[j][num] = false;bucket[bid][num] = false;board[i][j] = '.';}}}// 如果尝试完了1~9也没有返回true，说明没有符合条件的解，直接返回falsereturn false;}
}

三、解题思路 

一个一个位置玩DFS暴力尝试。将所有可能填入的数字都尝试一边，找到符合条件的答案就返回。题目保证每一个输入只有唯一的一个解。