• 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

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