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。