2.1.14 Valid Sudoku

Link: https://oj.leetcode.com/problems/valid-sudoku/

這題簡單，基本一次過。不用再做。

Time: O(3*n^2), n=9

注意：循環時用：for(int j = 0; j < board[0].length; j++)，不要用for(int j = 0; j < board[i].length; j++){

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        //check rows
        for(int i = 0; i < board.length; i++){
            HashSet<Character> set = new HashSet<Character>();
            for(int j = 0; j < board[0].length; j++){
                if(set.contains(board[i][j])){
                    return false;
                }
                if(board[i][j] != '.'){
                    set.add(board[i][j]);
                }
            }
        }
        //check columns
        for(int j = 0; j < board[0].length; j++){
            HashSet<Character> set = new HashSet<Character>();
            for(int i = 0; i < board.length; i++){
                if(set.contains(board[i][j])){
                    return false;
                }
                 if(board[i][j] != '.'){
                    set.add(board[i][j]);
                }
            }
        }
        //check squares
        for(int m = 0; m < 3; m++){
            for(int n = 0; n < 3; n++){
                HashSet<Character> set = new HashSet<Character>();
                for(int i = 0; i < 3; i++){
                    for(int j = 0; j < 3; j++){
                        if(set.contains(board[3*m+i][3*n+j])){
                            return false;
                        }
                        if(board[3*m+i][3*n+j] != '.'){
                            set.add(board[3*m+i][3*n+j]);
                        }
                    }
                }
            }
        }
        return true;
    }
}

更簡單的寫法可以寫成：

if(board[i][j]!='.' && !hashSet.add(board[i][j])){//if board[i][j] exist
<span style="white-space:pre">	</span>return false;
}

相關題目：Sudoku Solver