Leetcode-36-有效的数独

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。

image.png
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。

---

分析：

题意很容易理解，是一道关于二维数组的题目。只需要满足行、列、九宫格都没有重复的数字就可以返回true，只要有一个地方不满足，返回false即可。
行和列的判断很简单，很容易就可以搞定。
比较难的是九宫格的判断，当然如果用穷举的方法也可以实现，但是让代码写的优雅也是我们的目标。

接下来给出java解答：

解法一：

使用HashSet的不能重复添加相同元素的特性来实现，首先是行和列的判断，接下来针对九宫格创建了9个HashSet，分别盛放9个九宫格的元素，这样的方法很容易理解，但不够优雅，请看解法二。

    public boolean isValidSudoku(char[][] board) {
        Set<Character> rowSet = new HashSet<>();
        Set<Character> lineSet = new HashSet<>();

        Set<Character> set1 = new HashSet<>();
        Set<Character> set2 = new HashSet<>();
        Set<Character> set3 = new HashSet<>();
        Set<Character> set4 = new HashSet<>();
        Set<Character> set5 = new HashSet<>();
        Set<Character> set6 = new HashSet<>();
        Set<Character> set7 = new HashSet<>();
        Set<Character> set8 = new HashSet<>();
        Set<Character> set9 = new HashSet<>();

        for (int i = 0; i < 9; i++) {
            rowSet.clear();
            lineSet.clear();
            for (int j = 0; j < 9; j++) {
                if (board[j][i] != '.') {
                    if (!lineSet.add(board[j][i])) {
                        return false;
                    }
                }

                if (board[i][j] == '.') {
                    continue;
                }

                if (!rowSet.add(board[i][j])) {
                    return false;
                }

                if (j < 3) {
                    if (i < 3) {
                        if (!set1.add(board[i][j])) {
                            return false;
                        }
                    }

                    if (i >= 3 && i < 6) {
                        if (!set4.add(board[i][j])) {
                            return false;
                        }
                    }

                    if (i >= 6) {
                        if (!set7.add(board[i][j])) {
                            return false;
                        }
                    }
                }

                if (j >= 3 && j < 6) {
                    if (i < 3) {
                        if (!set2.add(board[i][j])) {
                            return false;
                        }
                    }

                    if (i >= 3 && i < 6) {
                        if (!set5.add(board[i][j])) {
                            return false;
                        }
                    }

                    if (i >= 6) {
                        if (!set8.add(board[i][j])) {
                            return false;
                        }
                    }
                }

                if (j >= 6) {
                    if (i < 3) {
                        if (!set3.add(board[i][j])) {
                            return false;
                        }
                    }

                    if (i >= 3 && i < 6) {
                        if (!set6.add(board[i][j])) {
                            return false;
                        }
                    }

                    if (i >= 6) {
                        if (!set9.add(board[i][j])) {
                            return false;
                        }
                    }
                }
            }
        }
        return true;
    }

解法二：

这种方法的基本思路与上一种解法相同，不同的是，不需要创建9个Set对象来盛放九宫格的元素，只需要一个就够了。代码中的x和y是依靠一定的计算方法将当前遍历的i和j转换成九宫格的二维数组坐标。如此就显得非常优雅。

    public boolean isValidSudoku(char[][] board) {
        Set<Character> rowSet = new HashSet<>();
        Set<Character> lineSet = new HashSet<>();
        Set<Character> sudokuSet = new HashSet<>();

        int k = 0;
        int x, y;

        for (int i = 0; i < 9; i++) {
            rowSet.clear();
            lineSet.clear();
            sudokuSet.clear();
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.' && !rowSet.add(board[i][j])) {
                    return false;
                }

                if (board[j][i] != '.' && !lineSet.add(board[j][i])) {
                    return false;
                }

                x = (i / 3) * 3 + (k % 9) / 3;
                y = j % 3 + ((k / 9) % 3) * 3;

                if (board[x][y] != '.' && !sudokuSet.add(board[x][y])) {
                    return false;
                }
                k += 1;
            }
        }
        return true;
    }