哈希表//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。

  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。
class Solution {
    public boolean isValidSudoku(char[][] board) {
        for(int i = 0; i < 9; i++){
            HashSet<Character> line = new HashSet<>();
            HashSet<Character> col = new HashSet<>();
            HashSet<Character> cube = new HashSet<>();
            for(int j = 0; j < 9; j++){
                if(board[i][j] != '.'&&!line.add(board[i][j]))
                    return false;
                if(board[j][i] != '.'&&!col.add(board[j][i]))
                    return false;
                int m = i/3*3+j/3;
                int n = i%3*3+j%3;
                if(board[m][n] != '.'&& !cube.add(board[m][n]))
                    return false;
            }
        }
        return true;
    }
}

 

class Solution {
    public boolean isValidSudoku(char[][] board) {
        for(int i = 0; i < 9; i++){
            int[] bit_map_row = new int[9];
            int[] bit_map_col = new int[9];
            int[] bit_map_cube = new int[9];
            for(int j = 0; j < 9; j++){
                if(board[i][j] != '.')
                    if(bit_map_row[board[i][j]-'1'] == 1)
                        return false;
                    else
                        bit_map_row[board[i][j]-'1'] = 1;
                if(board[j][i] != '.')
                    if(bit_map_col[board[j][i]-'1'] == 1) 
                        return false;
                    else
                        bit_map_col[board[j][i]-'1'] = 1;
                int m = 3*(i/3)+j/3;
                int n = 3*(i%3)+j%3;
                int val = board[m][n];
                if(val != '.')
                    if(bit_map_cube[val-'1'] == 1)
                        return false;
                    else
                        bit_map_cube[val-'1'] = 1;
                        
            }
            
        }
        return true;
            
    }
}

 

posted @ 2019-01-04 19:18  strawqqhat  阅读(167)  评论(0编辑  收藏  举报
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