LeetCode 精选 TOP 面试题（Java 实现）—— 有效的数独

一、题目描述

1.1 题目

• 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。

• 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。

• 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。

• 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。

1.2 知识点

• 数组

1.3 题目链接

• https://leetcode-cn.com/problems/valid-sudoku/

二、解题思路

2.1 自研思路

  思路比较简单，官方给的解法上面使用了哈希表来进行处理，但其实没有必要，使用数组即可解决。而对于行、列和块中重复数字的检测，其实也可以在一次扫描中完成，即创建三个 9*9 的数组，来记录每一行、每一列和每一块中已出现过的数字情况，因为使用数组，所以用 O(1) 的时间复杂度即可完成数字的标记工作，然后只需在对每个数字进行标记之前先确认该数字有没有在该行、该列或该块中出现过即可，如果出现过即数独无效。

  这里需要注意的是对于块的处理，在这里使用 blockIndex = i / 3 * 3 + j / 3，即首先因为数独表格每一列存在三个块，所以首先将当前行下标除三求出其所在块在每一列的三个块中的位置，然后再将其乘三得到该块所在行的首块下标（每一行存在三个块），最后再加上列下标除三（每一行存在三个块）得到当前坐标所在的块在整体表格中的位置。

三、实现代码

3.1 自研实现

class Solution {
    public boolean isValidSudoku(char[][] board) {
        boolean[][] row = new boolean[9][9];
        boolean[][] col = new boolean[9][9];
        boolean[][] block = new boolean[9][9];

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