37. 解数独 - 力扣（LeetCode）-Toy模板网

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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：

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。

数独部分空格内已填入了数字，空白格用 ‘.’ 表示。
37. 解数独 - 力扣（LeetCode）,leetcode,算法,职场和发展

输入示例

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"]]

输出示例

[["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"]]

解题思路
37. 解数独 - 力扣（LeetCode）,leetcode,算法,职场和发展

解题代码文章来源地址https://www.toymoban.com/news/detail-820484.html

class Solution {
    public void solveSudoku(char[][] board) {
        backtrack(board);
    }
    
    public boolean backtrack(char[][] board) {
        for(int i = 0; i < board.length; i++) {
            for(int j = 0; j < board[0].length; j++) {
                // 跳过原始数字
                if(board[i][j] != '.') {
                    continue;
                }
                // 只处理空白格，填写1-9
                for(char k = '1'; k <= '9'; k++) {
                    if(isValid(i, j, k, board)) {
                        board[i][j] = k;
                        boolean result = backtrack(board);
                        if(result) {
                            return true;
                        }
                        board[i][j] = '.';
                    }
                }
                // 九个是尝试完了还是不行，就返回 false
                return false;
            }
        }
        // 遍历完没有返回 false，说明找到合适棋盘位置了
        return true;
    }

    public boolean isValid(int row, int col, char k, char[][] board) {
        // 判断行是否重复
        for(int i = 0; i < 9; i++) {
            if(board[row][i] == k) {
                return false;
            }
        }
        // 判断列是否重复
        for(int j = 0; j < 9; j++) {
            if(board[j][col] == k) {
                return false;
            }
        }
        // 判断九宫格是否重复
        int startRow = (row / 3) * 3;
        int startCol = (col / 3) * 3;
        for (int i = startRow; i < startRow + 3; i++) { // 判断9方格里是否重复
            for (int j = startCol; j < startCol + 3; j++) {
                if (board[i][j] == k) {
                    return false;
                }
            }
        }
        return true;
    }
}