题目

判断一个 9x9 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。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 . 表示。

示例 1:

输入:
[
  ["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:
[
  ["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。

说明:

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。
给定数独序列只包含数字 1-9 和字符 .。
给定数独永远是 9x9 形式的。

方案一

func isValidSudoku(board [][]byte) bool {
    var m = map[byte]bool{} // 存放一行
    var n = [9]map[byte]bool{} // 对应的存放列
    var pos = map[[2]int]bool{
        [2]int{1, 1}: true, [2]int{1, 4}: true, [2]int{1, 7}: true,
        [2]int{4, 1}: true, [2]int{4, 4}: true, [2]int{4, 7}: true,
        [2]int{7, 1}: true, [2]int{7, 4}: true, [2]int{7, 7}: true,
    }

    for i, _ := range board {        
        for j, v := range board[i] {
            // 处理 3*3
            if _, ok := pos[[2]int{i, j}]; ok {
                var tmp = map[byte]bool{}
                for _, p := range [9][2]int{
                    [2]int{i - 1, j - 1}, [2]int{i, j - 1}, [2]int{i + 1, j - 1},
                    [2]int{i - 1, j}, [2]int{i, j}, [2]int{i + 1, j},
                    [2]int{i - 1, j + 1}, [2]int{i, j + 1}, [2]int{i + 1, j + 1},
                }{
                    if board[p[0]][p[1]] == byte(46) {
                        continue
                    }
                    if _, ok := tmp[board[p[0]][p[1]]]; ok {
                        return false
                    }
                    tmp[board[p[0]][p[1]]] = true
                }
            }

            if v == byte(46) {
                continue
            }
            if _, ok := m[v]; ok { // 行存在重复
                return false
            }
            m[v] = true

            // 处理列
            if n[j] == nil {
                n[j] = map[byte]bool{}
            }
            if _, ok := n[j][v]; ok { // 列存在重复
                return false
            }
            n[j][v] = true
        }
        m = map[byte]bool{} // 循环完一行，重新初始化
    }

    return true
}

上述方案没有什么特殊的，使用 map 分别对 行、列、3*3范围进行处理；空间复杂度较高

原文

https://leetcode-cn.com/explore/learn/card/hash-table/206/practical-application-design-the-key/822/

官方题解

https://leetcode-cn.com/problems/valid-sudoku/solution/you-xiao-de-shu-du-by-leetcode/