簡單來說就是為給一個二維陣列，請確定給的二維陣列是合法的數獨矩陣。

範例：

組合檢查

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        """
        Let's keep tracking each element and see where they belong
        """
        
        # Initialize tackers for each row, column and box
        row = {i: [] for i in range(9)}
        col = {i: [] for i in range(9)}
        box = {i: [] for i in range(9)}
        
        
        for row_idx in range(9):
            for col_idx in range(9):
                e = board[row_idx][col_idx]
                box_idx = (row_idx // 3) * 3 
                box_idx += col_idx // 3
                
                if e == '.':
                    continue
                    
                if e in row[row_idx] or e in col[col_idx] or e in box[box_idx]:
                    return False
                else:
                    row[row_idx].append(e)
                    col[col_idx].append(e)
                    box[box_idx].append(e)
        return True  

這邊主要是我們會有三個檢查的分組，首先我們先產生該分組：

row: {0: [], 1: [], 2: [], 3: [], 4: [], 5: [], 6: [], 7: [], 8: []}
col: {0: [], 1: [], 2: [], 3: [], 4: [], 5: [], 6: [], 7: [], 8: []}
box: {0: [], 1: [], 2: [], 3: [], 4: [], 5: [], 6: [], 7: [], 8: []}

基本上裡面各個list之中不能有重複的數字：

所以我們只要能在遞迴時，知道每個數字應該在哪些分組，即可完成。

橫的跟直的都非常容易離解，如拆開來看會像是：

for row_idx in range(9):
    for col_idx in range(9):
        e = board[row_idx][col_idx]
        if e == '.':
            continue

        if e in row[row_idx]:
            return False
        else:
            row[row_idx].append(e)

for row_idx in range(9):
    for col_idx in range(9):
        e = board[row_idx][col_idx]
        if e == '.':
            continue

        if e in col[col_idx]:
            return False
        else:
            col[col_idx].append(e)

箱型檢查則是使用了位置的idx進行分組，如拆開來看會像是：

for row_idx in range(9):
    for col_idx in range(9):
        e = board[row_idx][col_idx]
        box_idx = (row_idx // 3) * 3 
        box_idx += col_idx // 3

        if e == '.':
            continue

        if e in box[box_idx]:
            return False
        else:
            box[box_idx].append(e)
return True  

位置的idx會是這樣

1.

0 + 0 = 0
0 + 0 = 0
0 + 0 = 0
0 + 1 = 1
0 + 1 = 1
0 + 1 = 1
0 + 2 = 2
0 + 2 = 2
0 + 2 = 2
0 + 0 = 0
0 + 0 = 0
0 + 0 = 0
0 + 1 = 1
0 + 1 = 1
0 + 1 = 1
0 + 2 = 2
0 + 2 = 2
0 + 2 = 2
0 + 0 = 0
0 + 0 = 0
0 + 0 = 0
0 + 1 = 1
0 + 1 = 1
0 + 1 = 1
0 + 2 = 2
0 + 2 = 2
0 + 2 = 2

2.

3 + 0 = 3
3 + 0 = 3
3 + 0 = 3
3 + 1 = 4
3 + 1 = 4
3 + 1 = 4
3 + 2 = 5
3 + 2 = 5
3 + 2 = 5
3 + 0 = 3
3 + 0 = 3
3 + 0 = 3
3 + 1 = 4
3 + 1 = 4
3 + 1 = 4
3 + 2 = 5
3 + 2 = 5
3 + 2 = 5
3 + 0 = 3
3 + 0 = 3
3 + 0 = 3
3 + 1 = 4
3 + 1 = 4
3 + 1 = 4
3 + 2 = 5
3 + 2 = 5
3 + 2 = 5

3.

6 + 0 = 6
6 + 0 = 6
6 + 0 = 6
6 + 1 = 7
6 + 1 = 7
6 + 1 = 7
6 + 2 = 8
6 + 2 = 8
6 + 2 = 8
6 + 0 = 6
6 + 0 = 6
6 + 0 = 6
6 + 1 = 7
6 + 1 = 7
6 + 1 = 7
6 + 2 = 8
6 + 2 = 8
6 + 2 = 8
6 + 0 = 6
6 + 0 = 6
6 + 0 = 6
6 + 1 = 7
6 + 1 = 7
6 + 1 = 7
6 + 2 = 8
6 + 2 = 8
6 + 2 = 8

如果覺得合在一起難以理解，可以參考