36.有效的數獨-Python-LeetCode
一、題目
判斷一個 9x9 的數獨是否有效。只需要根據以下規則,驗證已經填入的數字是否有效即可。
- 數字 1-9 在每一行只能出現一次。
- 數字 1-9 在每一列只能出現一次。
- 數字 1-9 在每一個以粗實線分隔的 3x3 宮內只能出現一次。
數獨部分空格內已填入了數字,空白格用 ‘.’ 表示。 例1:
輸入: [ [“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”] ] 輸出: true
例2:
輸入: [ [“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”] ] 輸出: false
解釋: 除了第一行的第一個數字從 5 改為 8 以外,空格內其他數字均與 示例1 相同。 但由於位於左上角的 3x3 宮內有兩個 8 存在, 因此這個數獨是無效的。
二、解法
- 先建立三個空陣列
row
、col
、cell
,以cell
為例,裡面的每個空字典都代表一個 3×3單元格,然後我們需要把資料一個個填進去 - 遍歷整個二維陣列,然後邊遍歷邊把陣列分別存入到 行
row
, 列col
, 3×3單元格cell
內的字典,存為key
,而不是value
。 - 然後我們就可以判斷,行、列、3×3單元格 對應的字典內是否已經存在
board[x][y]
這個鍵名,如果存在,那麼說明重複了,返回False
- 注意,字典中的值這裡都為1,但是沒有任何意義,你可以隨意更改
- 把陣列存入 3×3的單元格是一個難點,num = 3*(x//3)+y//3,這個式子是關鍵,可以找個數獨,然後代入進去好好理解下
- 當然你也可以不用這個式子,用
if/else
語句來判斷也行,那樣比較好理解,但是不如這個式子簡潔 - 類似於:
if y<3 : ... elif 3<=y<6 : ... elif 6<=y : ...
,
程式碼如下:
#row,col,cell分別代表行,列,3x3單元格
row, col, cell =
[{}, {}, {}, {}, {}, {}, {}, {}, {}],
[{}, {}, {}, {}, {}, {}, {}, {}, {}],
[{}, {}, {}, {}, {}, {}, {}, {}, {}]
for x in range(9):
for y in range(9):
#取得單元格
num = 3*(x//3)+y//3
temp = board[x][y]
#不需要存入 '.'
if temp != '.':
if (temp not in row[x]
and temp not in col[y]
and temp not in cell[num]):
row[x][temp] = '1'
col[y][temp] = '1'
cell[num][temp] = '1'
else:
return False
return True
時間 64ms,擊敗了 99.3%
結語
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