36. 有效的数独

36. 有效的数独(难度:中等)

题目链接:https://leetcode-cn.com/problems/valid-sudoku/

题目描述:

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。

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。

解法一:

我们可以分别用三个二维数组来标记每一个数独数字在每一行、每一列、每一个以粗实线分隔的 3x3 宫内是否出现过,我们用二维数组的第一维来表示每一组的组数,三种情况都是9组。

若使用num = board[i][i]来遍历,若num != '.',那么

  • 行级数独的标记就是row[i][num]
  • 列级数独的标记就是col[j][num]
  • 块级数独的标记就是lump[lumpNum][num],其中lumpNum表示块的序号,从左到右,从上到下分成9块,lumpNum=i / 3 * 3 + j / 3

我们每遍历一个数独,只需要判断该数独在上述的三种情况是否重复出现即可。

代码:

class Solution {
    public boolean isValidSudoku(char[][] board) {
        // 记录每一行的数独
        boolean[][] row = new boolean[9][9];
        // 记录每一列的数独
        boolean[][] col = new boolean[9][9];
        // 记录每一方块的数据
        boolean[][] lump = new boolean[9][9];

        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char num = board[i][j];
                if (num != '.') {
                    num -= '1';
                    int lumpNum = i / 3 * 3 + j / 3;
                    if (row[i][num] || col[j][num] || lump[lumpNum][num]) {
                        return false;
                    } else {
                        row[i][num] = true;
                        col[j][num] = true;
                        lump[lumpNum][num] = true;
                    }
                }
            }
        }
        return true;
    }
}
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