Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

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

A partially filled sudoku which is valid.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        HashSet<Character> ht = new HashSet <Character>();
        for(int i=0;i<9;i++){
            for(int j=0;j<9;j++){
                if(board[i][j]!='.'){
                    if(ht.contains(board[i][j])){
                    	//System.out.println('1');
                        return false;
                    }else{
                        ht.add(board[i][j]);
                    }
                    
                }
            }
            ht.clear();
            //System.out.println(ht.isEmpty());
            for(int j=0;j<9;j++){
                if(board[j][i]!='.'){
                    if(ht.contains(board[j][i])){
                    	//System.out.println('2');
                        return false;
                    }else{
                        ht.add(board[j][i]);
                    }
                    
                }
            }
            ht.clear();
           // System.out.println("here");
            
        }
        for(int n=0;n<3;n++){
            for(int m=0;m<3;m++){
                for(int i=0;i<3;i++){
                    for(int j=0;j<3;j++){
                        if(board[i+n*3][j+m*3]!='.'){
                            if(ht.contains(board[i+n*3][j+m*3])){
                            	//System.out.println('3');
                                return false;
                            }else{
                                ht.add(board[i+n*3][j+m*3]);
                            }
                        }   
                    }
                   
                }
                 ht.clear();
            }
        }
        return true;
    }
}

答案写的比我的简洁好多

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        int n = board.length;
        boolean[][] row = new boolean[n][n];
        boolean[][] col = new boolean[n][n];
        boolean[][] block = new boolean[n][n];
        for(int i = 0; i < n; i++) {
        	for(int j = 0; j < n; j++) {
        		if(board[i][j] == '.') continue;
        		int c = board[i][j] - '1';
        		if(row[i][c] || col[j][c] || block[i/3*3 + j/3][c]) return false;
        		row[i][c] = col[j][c] = block[i/3*3 + j/3][c] = true; 
        	}
        }
        return true;
    }
}

这种方法可以一次性对于三种requirements全都进行判断。时间复杂度减少三分之一，空间复杂度加大。

这里要主要考虑对于每一个block，如何表示出来。答案中也是用每一行表示一个block。