36. Valid Sudoku

Q:
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.

A:   就是道简单的实现题

class Solution:
    def isValidSudoku(self, board: list[list[str]]) -> bool:
        def helper(row: int, col: int, board: list[list[str]]) -> bool:
            ch, count = board[row][col], 0 
            if ch == '.':
                return True
            for j in range(9):
                if board[row][j] == ch:
                    count += 1
            if count > 1:
                return False
            count = 0
            for i in range(9):
                if board[i][col] == ch:
                    count += 1
            if count > 1:
                return False
            r, c = row // 3 * 3, col // 3 * 3
            count = 0
            for i in range(r, r+3):
                for j in range(c , c+3):
                    if board[i][j] == ch:
                        count += 1
            return count <= 1

        for row in range(9):
            for col in range(9):
                if not helper(row, col, board):
                    return False
        return True







public class Solution {
    public boolean isValidSudoku(char[][] board) {
        // if(board==null || board.length ==0 || board.length !=9 || board[0].length!=9)
        //     return false;
        for(int i =0;i<board.length;i++){
            for(int j =0;j< board[0].length;j++)
                if( !isValid(i,j,board))
                    return false;
        }
        return true;
    }
    private boolean isValid(int row, int col , char[][] board){
        char ch = board[row][col];
        if(ch == '.')
            return true;
        for(int i = row+1; i<board.length;i++)
            if(board[i][col] == ch)
                return false;
        for(int j = col+1;j<board[0].length;j++)
            if(board[row][j] == ch)
                return false;
        board[row][col] = '.';// temporaryly change it 
        for(int i = 0;i<3;i++)
            for(int j = 0;j<3;j++)
                if(board[row/3 * 3 +i][col/3*3+j] == ch)
                    return false;
        board[row][col]=ch;
        return true;
    }
}

这里，在小方框里，忘记check是否是原来的位置。

class Solution:

    def isValidSudoku(self, board: list[list[str]]) -> bool:

        def checkSet( arr: list[str]) -> bool:

            S = set()

            for ch in arr:

                if ch == '.':

                    continue

                if ch in S:

                    return False

                S.add(ch)

            return True

        # check each row

        for row in board:

            if not checkSet(row):

                return False

        # check each column

        for j in range(9):

            if not checkSet([row[j] for row in board]):

                return False

        # check 3 X 3

        for row in range(0,9,3):

            for col in range(0,9,3):

                B = [board[r][c]

for r in range(row, row+3)

for c in range(col, col+3)]

                if not checkSet([ch for row in B for ch in row]):

                    return False

        return True

public class ValidSudoku {
    public boolean isValidSudoku(char[][] board) {
        // we only check valid, but not solvable
        if(board==null || board.length ==0 || board.length !=9 || board[0].length!=9){
            return false;
        }

        // check valid row   & check valid chars
        boolean[] validArray;
        for (int i = 0; i < 9; i++) { // for each row
            validArray = new boolean[10];
            for (int j = 0; j < 9; j++) {
                // check valid of chars
                if (! (board[i][j] >= '0' && board[i][j] <= '9')
                        &&  board[i][j] != '.') {
                    return false;
                }
                if(board[i][j]=='.')continue;
                int value = Integer.valueOf("" + board[i][j]);
                if (validArray[value]) {
                    return false;
                } else {
                    validArray[value] = true;
                }
            }
        }
        // check valid col
        for (int j = 0; j < 9; j++) {
            validArray = new boolean[10];
            for (int i = 0; i < 9; i++) { // for each row
                if(board[i][j]=='.')continue;
                int value = Integer.valueOf("" + board[i][j]);
                if (validArray[value]) {
                    return false;
                } else {
                    validArray[value] = true;
                }
            }
        }
        // check valid square
        for (int outerI = 0; outerI < 3; outerI++) {
            for (int outerJ = 0; outerJ < 3; outerJ++) {
                validArray = new boolean[10];
                for (int i = 0; i < 3; i++) {
                    for (int j = 0; j < 3; j++) {
                        if(board[3*outerI+i][3*outerJ+j]=='.')continue;
                        int value = Integer.valueOf("" + board[3*outerI+i][3*outerJ+j]);
                        if (validArray[value]) {
                            return false;
                        } else {
                            validArray[value] = true;
                        }
                    }
                }
            }
        }
        return true;
    }
}

Mistakes:
1： 对每个board上的数字， 当转换到int类型的时候，忘了检查，是否为'.'   ， 造成了 Exception
2： 在检查小方格的时候， 忘了算上outerI，  outerJ的基数， 哎， SB啊
int value = Integer.valueOf("" + board[3*outerI+i][3*outerJ+j]);


Learned: