LeetCode--036--有效的数独(java)

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。

rows:看行是否有重复元素 cols:看列是否有重复元素 cube:看各个9*9是否满足题意。

int rowIndex = 3 * (i / 3);

int colIndex = 3 * (i % 3);

0 1 2 3 4 5 6 7 8

0 0 0 3 3 3 6 6 6  rowIndex

0 3 6 0 3 6 0 3 6 colIndex

0 0 0 1 1 1 2 2 2  j / 3

0 1 2 0 1 2 0 1 2  j % 3

 1 class Solution {
 2     public boolean isValidSudoku(char[][] board) {
 3         for(int i = 0;i < board.length;i++){
 4             HashSet<Character>  rows = new HashSet<>();
 5             HashSet<Character>  cols = new HashSet<>();
 6             HashSet<Character> cube = new HashSet<>();
 7             for(int j = 0;j < board[0].length;j++){
 8                 if(board[i][j] != '.' && !rows.add(board[i][j])) return false; //添加成功返回1
 9                 if(board[j][i] != '.' && !cols.add(board[j][i])) return false;
10                 
11                 int rowIndex = 3 * (i / 3);
12                 int colIndex = 3 * (i % 3);
13                 
14                 if(board[rowIndex + j / 3][colIndex + j % 3] != '.' && !cube.add(board[rowIndex + j / 3][colIndex + j % 3])) return false;
15             }
16         }
17         return true;
18     }
19 }

2019-04-29 19:04:12

原文地址:https://www.cnblogs.com/NPC-assange/p/10792305.html