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。

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.com/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。

func isValidSudoku(board [][]byte) bool {
    boardLength := len(board)
    boardExcept := 3
    for y := 0; y < boardLength; y++ {
        dx := map[byte]int{}
        dy := map[byte]int{}
        for x := 0; x < boardLength; x++ {
            valX := board[y][x]
            if valX != '.' {
                if _, ok := dx[valX]; ok {
                    return false
                } else {
                    dx[valX] = 1
                }
            }
            valY := board[x][y]
            if valY != '.' {
                if _, ok := dy[valY]; ok {
                    return false
                } else {
                    dy[valY] = 1
                }
            }
        }
    }

    for c1 := 0; c1 < boardExcept; c1++ {
        lz := [][]byte{}
        for y1 := 0; y1 < boardLength; y1++ {
            if y1/3 == c1 {
                lz = append(lz, board[y1])
            }
        }
        for c2 := 0; c2 < boardExcept; c2++ {
            dz := map[byte]int{}
            for c3 := 0; c3 < boardExcept; c3++ {
                for c4 := 0; c4 < boardLength; c4++ {
                    valZ := lz[c3][c4]
                    if c4/3 == c2 {
                        if valZ != '.' {
                            if _, ok := dz[valZ]; ok {
                                return false
                            } else {
                                dz[valZ] = 1
                            }
                        }
                    }
                }
            }
        }
        
    }

    return true
}

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        for y in range(9):
            lx = {}
            ly = {}
            for x in range(9):
                valX = board[y][x]
                if valX != ".":
                    if lx.get(valX) == None:
                        lx[valX] = 1
                    else:
                        return False
                valY = board[x][y]
                if valY != ".":
                    if ly.get(valY) == None:
                        ly[valY] = 1
                    else:
                        return False
        for j in range(3):
            lz1 = []
            for i in range(9):
                if i // 3 == j:
                    lz1.append(board[i])
            for m in range(3):
                lz2 = []
                for k in range(3):
                    for n in range(9):
                        if n // 3 == m and lz1[k][n] != ".":
                            if lz1[k][n] in lz2:
                                return False
                            else:
                                lz2.append(lz1[k][n])                        
        return True
                        

 结束！