有效的数独 python_Python判断有效的数独算法示例
本文實例講述了Python判斷有效的數獨算法。分享給大家供大家參考，具體如下：
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。
1. 數字 1-9 在每一行只能出現一次。
2. 數字 1-9 在每一列只能出現一次。
3. 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。
但由于位于左上角的 3x3 宮內有兩個 8 存在, 因此這個數獨是無效的。
二、解法
先創建三個空數組 row、col、cell，以 cell 為例，里面的每個空字典都代表一個 3×3單元格，然后我們需要把數據一個個填進去
遍歷整個二維數組，然后邊遍歷邊把數組分別存入到 行 row , 列 col , 3×3單元格 cell 內的字典，存為key ，而不是 value 。
然后我們就可以判斷,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，那么說明重復了，返回 False
注意，字典中的值這里都為1，但是沒有任何意義，你可以隨意更改
把數組存入 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，然后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
類似于: if y<3 : ... elif 3<=y<6 : ... elif 6<=y : ...,
代碼如下：
#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%
希望本文所述對大家Python程序設計有所幫助。
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