36.有效的数独

请你判断一个 9 x 9 的数独是否有效。只需要 根据以下规则 ，验证已经填入的数字是否有效即可。

  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）

注意：

  • 一个有效的数独（部分已被填充）不一定是可解的。
  • 只需要根据以上规则，验证已经填入的数字是否有效即可。
  • 空白格用 '.' 表示。

示例 1：

img

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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：true

Solution

能想到用int[][][] subboxes = new int[3][3][9]和subboxes[i / 3][j / 3][index]++，存储分块元素的人是天才

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class Solution {
    public boolean isValidSudoku(char[][] board) {
        // rows，columns和subboxes用于记录每行、每列和每个3x3子数独中数字的出现次数
        int[][] rows = new int[9][9];
        int[][] columns = new int[9][9];
        int[][][] subboxes = new int[3][3][9];
        
        // 遍历数独的每个格子
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char c = board[i][j]; // 获取当前格子的字符
                if (c != '.') { // 如果当前格子有数字
                    int index = c - '0' - 1; // 计算数字对应的索引（由于数字从1开始，数组索引从0开始，所以需要减1）
                    
                    // 在相应的位置增加计数
                    rows[i][index]++;
                    columns[j][index]++;
                    subboxes[i / 3][j / 3][index]++;
                    
                    // 检查当前数字在同一行、同一列或同一个子数独中是否已经出现过
                    if (rows[i][index] > 1 || columns[j][index] > 1 || subboxes[i / 3][j / 3][index] > 1) {
                        return false; // 如果任何计数大于1，则数独无效
                    }
                }
            }
        }
        return true; // 如果所有数字都满足数独的条件，则数独有效
    }
}