Leetcode36. 有效的數獨-python

難度：中等

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。

上圖是一個部分填充的有效的數獨。

數獨部分空格內已填入了數字，空白格用 '.' 表示。

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

示例 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 改爲 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。
說明:

一個有效的數獨（部分已被填充）不一定是可解的。
只需要根據以上規則，驗證已經填入的數字是否有效即可。
給定數獨序列只包含數字 1-9 和字符 '.' 。
給定數獨永遠是 9x9 形式的。

思路：這個地方我沒什麼好的思路，直接採取遍歷的方式，有兩種想法，第一種是申請27個數組，9個用來存入行判斷，9個判斷列，9個判斷3*3；只需要遍歷即可，每個數據都存入對於的數組內（一個數要存3個數組）然後中間判斷是否存在即可；第二種方法是，只申請一個數組，遍歷三遍，行、列、3*3各一次，每次用一個數組判斷就可以。

代碼如下（3次循環，時間複雜度打敗60%+，空間僅打敗6%+）：

class Solution(object):
    def isValidSudoku(self, board):
        """
        :type board: List[List[str]]
        :rtype: bool
        """
        l=len(board)
        #要麼遍歷一次，申請3*l個數組來判斷，也可以遍歷三次，只需要一個數組即可
        list1=[]
        flag=0
        #行判斷
        for i in range(l):
            list1=[]
            for j in range(l):
                if board[i][j] not in list1 and board[i][j]!=".":
                    list1.append(board[i][j])
                elif board[i][j] in list1:
                    flag=1
                    break
        #不符合行則直接結束
        if flag==1:
            return False
        #列判斷
        for i in range(l):
            list1=[]
            for j in range(l):
                if board[j][i] not in list1 and board[j][i]!=".":
                    list1.append(board[j][i])
                elif board[j][i] in list1:
                    flag=1
                    break
        #不符合列則直接結束
        if flag==1:
            return False
        #九宮格判斷
        #將9*9分解爲9個3*3，座標分別爲（0，1，2）*（0，1，2）
        for i in range(l/3):
            for j in range(l/3):
                list1=[]
                #print(j)
                #現在遍歷3*3
                for k in range(l/3):
                    for m in range(l/3):
                        #這裏改了改判斷方式，其實都一樣的效果
                        if board[i*3+k][j*3+m]!=".":    
                            if board[i*3+k][j*3+m] not in list1 :
                                list1.append(board[i*3+k][j*3+m])
                            else:
                                flag=1
                                break
        if flag==1:
            return False
        else:
            return True