题目描述：
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。
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
解题思路：
-
先校验横向和竖向是否有重复，对于任意一个位置的数字来说，只有这一行没有重复的，并且这一列也没有重复的，说明这个数横竖符合条件.
-
然后判断3*3的九宫格是否有重复
解法一：
function isValidSudoku(board) {// 用来存放行的maplet rowMap = new Map();// 用来存放列的maplet columnMap = new Map();// 存放返回的结果let res = true;// 先校验横向和竖向是否有重复，对于任意一个位置的数字来说，只有这一行没有重复的，//并且这一列也没有重复的，说明这个数横竖符合条件// 所以要做的就是判断每一行每一列是否有重复的数字，这里行列一起判断，别浪费循环for (let i = 0; i < 9; i++) {for (let j = 0; j < 9; j++) {if (rowMap.has(board[i][j])) {res = false;break;} else {if (board[i][j] !== '.') {rowMap.set(board[i][j], 1);}}if (columnMap.has(board[j][i])) {res = false;break;} else {if (board[j][i] !== '.') {columnMap.set(board[j][i], 1);}}}// 每一行列判断完以后要清空两个maprowMap.clear();columnMap.clear();}// 如果行列判断完已经有重复了，就不用判断后面的九格内不重复了if (!res) {return res;}// 三行为一轮循环，这两个变量记录第几次行/列的跳跃，先把行的三次判断完，// 然后换下一组9个的let rown = 0;let coln = 0;// 然后检查九宫格是否有重复while (rown < 3 && coln < 3) {for (let i = 0; i < 3; i++) {for (let j = 0; j < 3; j++) {if (rowMap.has(board[rown * 3 + i][coln * 3 + j])) {res = false;break;} else {if (board[rown * 3 + i][coln * 3 + j] !== '.') {rowMap.set(board[rown * 3 + i][coln * 3 + j], 1);} }}}rown = rown + 1;rowMap.clear();if (rown === 3) {coln = coln + 1;rown = 0;}}return res;
};用时：
//507/507 cases passed (76 ms)
// Your runtime beats 84.31 % of typescript submissions
// Your memory usage beats 91.5 % of typescript submissions (44.3 MB)
)
