leetcode算法题-有效的数独

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题目

本题为leetcode探索初级算法中数组章节的一题

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。

1.数字 1-9 在每一行只能出现一次。
2.数字 1-9 在每一列只能出现一次。
3.数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

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:

1.一个有效的数独（部分已被填充）不一定是可解的。
2.只需要根据以上规则，验证已经填入的数字是否有效即可。
3.给定数独序列只包含数字 1-9 和字符 ‘.’ 。
4.给定数独永远是 9x9 形式的。

作者：力扣 (LeetCode)
链接：https://leetcode-cn.com/leetbook/read/top-interview-questions-easy/x2f9gg/
来源：力扣（LeetCode）
著作权归作者所有。商业转载请联系作者获得授权，非商业转载请注明出处。

我的解法

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class Solution {
    public boolean isValidSudoku(char[][] board) {
        for (int i = 0; i < 9; i++) {
            Set set = new HashSet();
            for (int j = 0; j < 9; j++) {
                if(board[i][j] != '.' && set.contains(board[i][j])){
                    return false;
                }
                set.add(board[i][j]);
            }
        }
        for (int i = 0; i < 9; i++) {
            Set set = new HashSet();
            for (int j = 0; j < 9; j++) {
                if(board[j][i] != '.' && set.contains(board[j][i])){
                    return false;
                }
                set.add(board[j][i]);
            }
        }
        return check9(board, 1, 1) && check9(board, 1, 4) && check9(board, 1, 7)
            && check9(board, 4, 1) && check9(board, 4, 4) && check9(board, 4, 7)
            && check9(board, 7, 1) && check9(board, 7, 4) && check9(board, 7, 7);
    }
    
    
   private static boolean check9(char[][] board, int center1, int center2) {
         Set set = new HashSet();
        if (board[center1 - 1][center2 - 1] != '.' && set.contains(board[center1 - 1][center2 - 1])) {
            return false;
        }
        set.add(board[center1 - 1][center2 - 1]);

        if (board[center1 - 1][center2] != '.' && set.contains(board[center1 - 1][center2])) {
            return false;
        }
        set.add(board[center1 - 1][center2]);

        if (board[center1 - 1][center2 + 1] != '.' && set.contains(board[center1 - 1][center2 + 1])) {
            return false;
        }
        set.add(board[center1 - 1][center2 + 1]);

        if (board[center1][center2 - 1] != '.' && set.contains(board[center1][center2 - 1])) {
            return false;
        }
        set.add(board[center1][center2 - 1]);

        if (board[center1][center2] != '.' && set.contains(board[center1][center2])) {
            return false;
        }
        set.add(board[center1][center2]);

        if (board[center1][center2 + 1] != '.' && set.contains(board[center1][center2 + 1])) {
            return false;
        }
        set.add(board[center1][center2 + 1]);

        if (board[center1 + 1][center2 - 1] != '.' && set.contains(board[center1 + 1][center2 - 1])) {
            return false;
        }
        set.add(board[center1 + 1][center2 - 1]);

        if (board[center1 + 1][center2] != '.' && set.contains(board[center1+1][center2])) {
            return false;
        }
        set.add(board[center1 + 1][center2]);

        if (board[center1 + 1][center2 + 1] != '.' && set.contains(board[center1 + 1][center2 + 1])) {
            return false;
        }
        return true;
    }
    
}

比较蠢的解法。代码比较臃肿。

官方解法

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class Solution {
  public boolean isValidSudoku(char[][] board) {
    // init data
    HashMap<Integer, Integer> [] rows = new HashMap[9];
    HashMap<Integer, Integer> [] columns = new HashMap[9];
    HashMap<Integer, Integer> [] boxes = new HashMap[9];
    for (int i = 0; i < 9; i++) {
      rows[i] = new HashMap<Integer, Integer>();
      columns[i] = new HashMap<Integer, Integer>();
      boxes[i] = new HashMap<Integer, Integer>();
    }

    // validate a board
    for (int i = 0; i < 9; i++) {
      for (int j = 0; j < 9; j++) {
        char num = board[i][j];
        if (num != '.') {
          int n = (int)num;
          int box_index = (i / 3 ) * 3 + j / 3;

          // keep the current cell value
          rows[i].put(n, rows[i].getOrDefault(n, 0) + 1);
          columns[j].put(n, columns[j].getOrDefault(n, 0) + 1);
          boxes[box_index].put(n, boxes[box_index].getOrDefault(n, 0) + 1);

          // check if this value has been already seen before
          if (rows[i].get(n) > 1 || columns[j].get(n) > 1 || boxes[box_index].get(n) > 1)
            return false;
        }
      }
    }

    return true;
  }
}

时间复杂度：O(1)O(1)，因为我们只对 81 个单元格进行了一次迭代。
空间复杂度：O(1)O(1)。