【求知=>算法】有效的数独

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【求知=>算法】有效的数独

请你判断一个9 x 9 的数独是否有效。只需要 根据以下规则 ,验证已经填入的数字是否有效即可。

数字1-9在每一行只能出现一次。 数字1-9在每一列只能出现一次。 数字1-9在每一个以粗实线分隔的 3x3 宫内只能出现一次。(请参考示例图)

注意:

  • 一个有效的数独(部分已被填充)不一定是可解的。
  • 只需要根据以上规则,验证已经填入的数字是否有效即可。
  • 空白格用 ’.’ 表示。

示例 1:

image.png

输入: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"]]
输出:true

示例 2:

输入: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"]]
输出:false
解释:除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

提示:

  • board.length == 9
  • board[I].length == 9
  • board[i][j] 是一位数字(1-9)或者 ’.’

解题思路

  1. 我们首先遍历9宫格的所有元素,然后使用3个二维数组遍历,记录对应的行,列以及3*3单元格是否有某个数字,如果出现冲突,直接返回false。
 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"]]
    
    
#     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"]]
    
    n = len(board)
    m = len(board[0])
    row = [[] * 9 for _ in range(9)] # 行
    col = [[] * 9 for _ in range(9)] # 列
    nine = [[] * 9 for _ in range(9)] #子数检查
    
    for i in range(n):
        for j in range(m):
            tmp = board[i][j]
            if not tmp.isdigit():
                continue
            if tmp in row[i]:
                return False
            if tmp in col[j]:
                return False
            if tmp in nine[(j // 3) * 3 + (i // 3)]:
                return False
            row[i].append(tmp)
            col[j].append(tmp)
            nine[(j // 3) * 3 + (i // 3)].append(tmp)
            
    return True
  1. python哈希表解法,记录每个数字的行号,列号,以及所在block的编号。
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"]]
        
        sudoku_dict = {}
        for i,row in enumerate(board):
            for j,item in enumerate(row):
                which_block = int(i/3)*3+int(j/3)
                if sudoku_dict.get(item) is not None:
                    if i in sudoku_dict[item]["row"] or j in sudoku_dict[item]["col"] or which_block in sudoku_dict[item]["block"]:
                        return False
                    else:
                        sudoku_dict[item]["row"].append(i)
                        sudoku_dict[item]["col"].append(j)
                        sudoku_dict[item]["block"].append(which_block)
                elif item!=".":
                    sudoku_dict[item] = {"row":[i],"col":[j],"block":[which_block]}
        return True

知识扩展


isdigit()方法的使用

博客来源:雨夜的博客