理解错题了 36. 有效的数独

 理解错题了，就是让你在矩阵中循环判断而已，我还想到了康托付展开

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。
 

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。
 

提示：

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

来源：力扣（LeetCode）
链接：https://leetcode.cn/problems/valid-sudoku
著作权归领扣网络所有。商业转载请联系官方授权，非商业转载请注明出处。

/**
 * @param {character[][]} board
 * @return {boolean}
 */
var isValidSudoku = function(board) {
    // 理解错题了，就是让你在矩阵中循环判断而已
    const rectArr=[]
    for(let i=0;i<9;i++){
        rectArr.push([i,i+1,0,9])
        rectArr.push([0,9,i,i+1])
    }
    for(let i=0;i<9;i=i+3){
        for(let j=0;j<9;j=j+3){
            rectArr.push([i,i+3,j,j+3])
        }
    }
    console.log(rectArr)
    let isOk=true;
    for(let i=0;i<rectArr.length;i++){
        if(!isOk){
            break;
        }
        const arr=[]
        const rect=rectArr[i];
        for(let j=rect[0];j<rect[1];j++){
            for(let k=rect[2];k<rect[3];k++){
                const t=board[j][k];
                if(t!=='.'){
                    if(arr.indexOf(t)===-1){
                        arr.push(t)
                    }else{
                        isOk=false;
                        break
                    }
                }
            }
        }
    }

    return isOk
};