【哈希-中等】36. 有效的数独

【题目】
判断一个 9x9 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。
上图是一个部分填充的有效的数独。
数独部分空格内已填入了数字，空白格用 ‘.’ 表示。
【示例 1】
输入:
[
[“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”]
]
输出: true
【示例 2】
输入:
[
[“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”]
]
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。
说明:
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。
给定数独序列只包含数字 1-9 和字符 ‘.’ 。
给定数独永远是 9x9 形式的。
【代码】
【Python】
【方法1：三次遍历】
三步走：
1、按行遍历
2、按列遍历
3、遍历九宫格（九个九宫格）

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        # 按行遍历
        for row in board:
            rest=[r for r in row if r!='.']
            if len(set(rest))!=len(rest):
                return False
        # 按列遍历
        for row in range(9):
            s=set()
            l=list()
            for col in range(9):
                if board[col][row]!='.':
                    s.add(board[col][row])
                    l.append(board[col][row])
            if len(s)!=len(l):
                return False
        # 遍历9个九宫格
        for row in range(3):
            for col in range(3):
                s=set()
                l=list()
                for i in range(3):
                    for j in range(3):
                        if board[i+row*3][j+col*3]!='.':
                            s.add(board[i+row*3][j+col*3])
                            l.append(board[i+row*3][j+col*3])
                if len(s)!=len(l):
                    return False
        return True

【方法二：dict】

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        cnt=dict()
        for i in range(9):
            for j in range(9):
                if board[i][j]!=".":
                    cnt[board[i][j]]=cnt.setdefault(board[i][j],[])
                    cnt[board[i][j]].append([i,j])
        for key in cnt:
            for i in range(len(cnt[key])):
                for j in range(i+1,len(cnt[key])):
                    if cnt[key][i][0]==cnt[key][j][0] or cnt[key][i][1]==cnt[key][j][1] or (cnt[key][i][0]//3==cnt[key][j][0]//3 and cnt[key][i][1]//3==cnt[key][j][1]//3):
                            print(key)
                            return False
        return True

【方法3：一次遍历】
执行用时：
48 ms, 在所有 Python3 提交中击败了81.02%的用户
内存消耗：
14.8 MB, 在所有 Python3 提交中击败了77.65%的用户

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        row,col,box=[[] for _ in range(9)],[[] for _ in range(9)],[[] for _ in range(9)]
        for i in range(9):
            for j in range(9):
                if board[i][j]!='.':
                    if board[i][j] in row[i] or board[i][j] in col[j] or board[i][j] in box[(i//3)*3+j//3]:
                        return False
                    else:
                        row[i].append(board[i][j])
                        col[j].append(board[i][j])
                        box[(i//3)*3+j//3].append(board[i][j])
        return True