36. Valid Sudoku 终极之牛逼方法

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

1. Each row must contain the digits ​​1-9​​ without repetition.
2. Each column must contain the digits ​​1-9​​ without repetition.
3. Each of the 9 ​​3x3​​ sub-boxes of the grid must contain the digits ​​1-9​​ without repetition.

The Sudoku board could be partially filled, where empty cells are filled with the character ​​'.'​​.

Example 1:

Input:
[
  ["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"]
]
Output: true

Example 2:

Input:
[
  ["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"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:

• A Sudoku board (partially filled) could be valid but is not necessarily solvable.
• Only the filled cells need to be validated according to the mentioned rules.
• The given board contain only digits ​​1-9​​ and the character ​​'.'​​.
• The given board size is always ​​9x9​​.

终极之无敌巧妙方法，理解不能

class Solution {
/*
　　int rowIndex = 3 * (i / 3);
　　int colIndex = 3 * (i % 3);
i~[0,8]
0 1 2 3 4 5 6 7 8
0 0 0 3 3 3 6 6 6 　　row Index
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
扫描cube时是从左到右从上到下扫描9个cube
*/
    public boolean isValidSudoku(char[][] board) {
       for(int i = 0; i < board.length; i++){
           HashSet<Character> rows = new HashSet<>();
            HashSet<Character> cols = new HashSet<>();
            HashSet<Character> cube = new HashSet<>();
           for(int j = 0; j < board[0].length; j++){
               if(board[i][j] != '.' && !rows.add(board[i][j])) return false;
               if(board[j][i] != '.' && !cols.add(board[j][i])) return false;
               
               int rowIndex = 3 * (i / 3);
               int colIndex = 3 * (i % 3);
               
               if(board[rowIndex + j / 3][colIndex + j % 3] != '.' && !cube.add(board[rowIndex + j / 3][colIndex + j % 3])) 
                   return false;
           }
       }
        return true;
    }
}

 好的 终于有一种看得懂的方法了，那就是遍历三次，检查三次

class Solution {
    public boolean isValidSudoku(char[][] board) {
        for(int i = 0; i < 9; i++) {
            if(!isvalidrow(board, i)) return false;
            if(!isvalidcol(board, i)) return false;
            for(int j = 0; j < 9; j++) {
                if(i % 3 == 0 && j % 3 == 0) {
                    if(!isvalids(board, i, j)) return false;
                }
            }                      
        }
        return true;
    }
    
    public boolean isvalidrow(char[][] ch, int r) {
        Set<Character> set = new HashSet();
        for(int i = 0; i < 9; i++) {
            if(ch[r][i] != '.') {
                if(!set.contains(ch[r][i])) set.add(ch[r][i]);
                else return false;
            }
        }
        return true;
    }
    
    public boolean isvalidcol(char[][] ch, int c) {
        Set<Character> set = new HashSet();
        for(int i = 0; i < 9; i++) {
            if(ch[i][c] != '.') {
                if(!set.contains(ch[i][c])) set.add(ch[i][c]);
                else return false;
            }
        }
        return true;
    }
    
        public boolean isvalids(char[][] ch, int r, int c) {
        Set<Character> set = new HashSet();
        for(int i = r; i < r + 3; i++) {
            for(int j = c; j < c + 3; j++) {
                if(ch[i][j] != '.') {
                if(!set.contains(ch[i][j])) set.add(ch[i][j]);
                else return false;
                }
            }
        }
        return true;
    }
}

 

 可不可以只遍历一次呢？可以，牛逼

class Solution {
public boolean isValidSudoku(char[][] board) {
    Set seen = new HashSet();
    for (int i=0; i<9; ++i) {
        for (int j=0; j<9; ++j) {
            char number = board[i][j];
            if (number != '.')
                if (!seen.add(number + " in row " + i) ||
                    !seen.add(number + " in column " + j) ||
                    !seen.add(number + " in block " + i/3 + "-" + j/3))
                    return false;
        }
    }
    return true;
}
}