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。

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。（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。

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.
Note:

A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.

提示：

board.length == 9
board[i].length == 9
board[i][j] 是一位数字（1-9）或者 ‘.’

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：true
示例 2：

输入：board =

[["8","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：false
解释：除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

解题：（没办法太菜了，只能把官方文档拿出来）

https://leetcode.cn/problems/valid-sudoku/solutions/1001859/you-xiao-de-shu-du-by-leetcode-solution-50m6/

方法一：一次遍历
有效的数独满足以下三个条件：

• 同一个数字在每一行只能出现一次；
• 同一个数字在每一列只能出现一次；
• 同一个数字在每一个小九宫格只能出现一次。

可以使用哈希表记录每一行、每一列和每一个小九宫格中，每个数字出现的次数。只需要遍历数独一次，在遍历的过程中更新哈希表中的计数，并判断是否满足有效的数独的条件即可。

对于数独的第 i 行第 j 列的单元格，其中 0≤i,j<9，该单元格所在的行下标和列下标分别为 i 和 j，该单元格所在的小九宫格的行数和列数分别为 [i/3]和[j/3]，其中0<=[i/3]，[j/3] < 3

由于数独中的数字范围是 1 到 9 ，因此可以使用数组代替哈希表进行计数。

具体做法是，创建二维数组 rows 和 columns分别记录数独的每一行和每一列中的每个数字的出现次数，创建三维数组 subboxes 记录数独的每一个小九宫格中的每个数字的出现次数，其中 rows[i][index]、columns[j][index]和 subboxes[i/3][j/3][index]分别表示数独的第 i 行第 j 列的单元格所在的行、列和小九宫格中，数字 index+1出现的次数，其中 0≤index<9，对应的数字 index+1满足 1≤index+1≤9。([i/3]，[j/3] 需要向下取整)。

如果 board[i][j]填入了数字 n，则将 rows[i][n−1]、columns[j][n−1]和 subboxes[i3][j3⌋][n−1]各加 1 。如果更新后的计数大于 1，则不符合有效的数独的条件，返回 false。

如果遍历结束之后没有出现计数大于 1 的情况，则符合有效的数独的条件，返回 true。

leetcode:https://leetcode.cn/problems/valid-sudoku/description/

var isValidSudoku = function (board) {
    const rows = new Array(9).fill(0).map(() => { return new Array(9).fill(0) })//9X9
    const columns = new Array(9).fill(0).map(() => { return new Array(9).fill(0) })//9X9
    const smallBoard = new Array(3).fill(0).map(() => { return new Array(3).fill(0).map(() => { return new Array(9).fill(0) }) })// 3X3X9
    //9X9
    for (let i = 0; i < 9; i++) {
        for (let j = 0; j < 9; j++) {
            const item = board[i][j]
            //找到不为空的数据
            if (item !== '.') {
            	//rows[i][index]、columns[j][index]和 subboxes[i/3][j/3][index]分别表示数独的第 i 行第 j 列的单元格所在的行、列和小九宫格中，数字 index+1出现的次数
                const index = item.charCodeAt() - "0".charCodeAt() - 1
                rows[i][index]++
                columns[j][index]++
                smallBoard[Math.floor(i / 3)][Math.floor(j / 3)][index]++
                if (rows[i][index] > 1 || columns[j][index] > 1 || smallBoard[Math.floor(i / 3)][Math.floor(j / 3)][index] > 1) {
                    return false
                }
            }
        }
    }
    return true
};

var isValidSudoku = function (board) {
    const rows = new Array(9).fill(0).map(() => { return new Array(9).fill(0) })
    const columns = new Array(9).fill(0).map(() => { return new Array(9).fill(0) })
    const smallBoard = new Array(3).fill(0).map(() => { return new Array(3).fill(0).map(() => { return new Array(9).fill(0) }) })
    for (let i = 0; i < 9; i++) {
        for (let j = 0; j < 9; j++) {
            const item = board[i][j]
            if (item !== '.') {
                const index = item.charCodeAt() - "0".charCodeAt() - 1
                rows[i][index]++
                columns[j][index]++
                smallBoard[Math.floor(i / 3)][Math.floor(j / 3)][index]++
                if (rows[i][index] > 1 || columns[j][index] > 1 || smallBoard[Math.floor(i / 3)][Math.floor(j / 3)][index] > 1) {
                    return false
                }
            }
        }
    }
    return true
};