[leet code] Valid Sudoku

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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.

This problem is strait forward, we need only implementing all the 3 rules of Sudoku.

The only thing we need to concern about is how to determine if numbers (in each rule) are duplicated. Accordingly, there are 2 approaches:

1. use HashSet.add to determine (if duplicate value added, method would return false)

2. use boolean array[265] (representing all the ASCII characters) to determine. Set boolean value true when new number (character) read. If the value has already true, then a duplicate value read.

Approach 1:

public class Solution {    public boolean isValidSudoku(char[][] board) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.                // rule1, column        for(int i=0; i<board[0].length; i++){            HashSet<Character> test = new HashSet<Character>();            for(int j=0; j<board.length; j++){                if(board[j][i]!='.' && !test.add(board[j][i])) return false;            }        }                // rule2, row        for(int i=0; i<board.length; i++){            HashSet<Character> test = new HashSet<Character>();            for(int j=0; j<board[0].length; j++){                if(board[i][j]!='.' && !test.add(board[i][j])) return false;            }        }                   // rule3, sub-box        for(int i=0; i<3; i++){            for(int j=0; j<3; j++){// for each sub-box                HashSet<Character> test = new HashSet<Character>();                for(int m=i*3; m<i*3+3; m++){//row                    for(int n=j*3; n<j*3+3; n++){//column                        if(board[m][n]!='.' && !test.add(board[m][n])) return false;                    }                }            }        }                return true;    }}

Approach2:

public class Solution {    public boolean isValidSudoku(char[][] board) {        // IMPORTANT: Please reset any member data you declared, as        // the same Solution instance will be reused for each test case.                // rule1, column        for(int i=0; i<board[0].length; i++){            boolean[] dupCheck = new boolean[256];            for(int j=0; j<board.length; j++){                if(board[j][i]!='.'){                    if(dupCheck[board[j][i]]==true) return false;                    else dupCheck[board[j][i]] = true;                                    }            }        }                // rule2, row        for(int i=0; i<board.length; i++){            boolean[] dupCheck = new boolean[256];            for(int j=0; j<board[0].length; j++){                if(board[i][j]!='.'){                    if(dupCheck[board[i][j]]==true) return false;                    else dupCheck[board[i][j]] = true;                                    }            }        }                   // rule3, sub-box        for(int i=0; i<3; i++){            for(int j=0; j<3; j++){// for each sub-box                boolean[] dupCheck = new boolean[256];                for(int m=i*3; m<i*3+3; m++){//row                    for(int n=j*3; n<j*3+3; n++){//column                        if(board[m][n] != '.'){                            if(dupCheck[board[m][n]]==true) return false;                            else dupCheck[board[m][n]] = true;                                                       }                    }                }            }        }                return true;    }}

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