LeetCode＃36. Effective Sudoku

subject :

Please judge a  9 x 9 Is the Sudoku effective . It only needs According to the following rules , Verify that the numbers you have filled are valid .

Numbers  1-9  Only once in a row .
Numbers  1-9  It can only appear once in each column .
Numbers  1-9  Separated by thick solid lines in each  3x3  Only once in the palace .（ Please refer to the example figure ）
 

Be careful ：

An effective Sudoku （ Part has been filled in ） Not necessarily solvable .
Just follow the above rules , Verify that the numbers you have filled are valid .
Blank space  '.'  Express .
 

Example 1：

Input ：board = 
[["5","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
Output ：true
Example 2：

Input ：board = 
[["8","3",".",".","7",".",".",".","."]
,["6",".",".","1","9","5",".",".","."]
,[".","9","8",".",".",".",".","6","."]
,["8",".",".",".","6",".",".",".","3"]
,["4",".",".","8",".","3",".",".","1"]
,["7",".",".",".","2",".",".",".","6"]
,[".","6",".",".",".",".","2","8","."]
,[".",".",".","4","1","9",".",".","5"]
,[".",".",".",".","8",".",".","7","9"]]
Output ：false
explain ： Except for the first number in the first line from 5 Change it to 8 outside , The other numbers in the space are the same as Example 1 identical . But because of the 3x3 There are two in the palace 8 There is , So this Sudoku is invalid .
 

Tips ：

board.length == 9
board[i].length == 9
board[i][j] It's a number （1-9） perhaps '.'

source ： Power button （LeetCode）
link ： Power button

Personally, I think this problem is mainly about the third condition of Sudoku judgment, which is relatively difficult , The first two conditions should be able to understand the principle by looking at the code , We mainly talk about the judgment process of the third condition .

Let's create a string first , The following methods will be used .

store="0123456789"

Conditions for a :

        for i in board:
            for j in store:
                num=i.count(j)
                if num>1:
                    return False

Condition 2 :

        nums=0
        while nums < 9:
            ls=[]
            for i in board:
                ls.append(i[nums])
            for j in store:
                if ls.count(j)>1:
                    return False
            nums+=1

Condition 3 :

We know that every Sudoku matrix has 9 A small cube matrix , So we can think of a way , Judge whether a small cube matrix is qualified each time , In this case, we have to judge 9 Time （ This is the cycle 9 Time ）.

Subdivide again , Every three small square matrices are in one row , So we can judge three times in each line （ loop 3 Time ）.

Through this , With the help of method 2 （ Create another list ） Same method , We can judge whether each of our small squares is qualified .

Here it is , I created two variables to help with the loop ,line: Indicates the number of rows indexed ; lis: Indicates the number of columns in the index .

        line=0;lis=0;ll=[]
        for i in range(3):  # Circle the three lines （ One big line corresponds to three small lines ）
            for i in range(3):  # Every big line （ Three small lines ）, Cycle three times ,3 ride 3 be equal to 9
                for j in board[line:line+3]:
                    for k in j[lis:lis+3]:
                        ll.append(k)
                for m in store:
                    if ll.count(m)>1:
                        return False
                lis+=3 # Advance the number of columns by three , Enter the next cube matrix of the big row 
                ll=[]  # Notice that after judging each small cube matrix , Clear list 
            lis=0    # After a large line of each cycle , Zero the number of columns 
            line+=3  # Go a long way 

  Last , If the above judgment does not return False, Then this is an effective Sudoku , We return at the end True that will do .

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        store="123456789"
        for i in board:
            for j in store:
                num=i.count(j)
                if num>1:
                    return False
        nums=0
        while nums < 9:
            ls=[]
            for i in board:
                ls.append(i[nums])
            for j in store:
                if ls.count(j)>1:
                    return False
            nums+=1
        line=0;lis=0;ll=[]
        for i in range(3):
            for i in range(3):
                for j in board[line:line+3]:
                    for k in j[lis:lis+3]:
                        ll.append(k)
                for m in store:
                    if ll.count(m)>1:
                        return False
                lis+=3 
                ll=[]
            lis=0
            line+=3
        return True