題目鏈接: https://leetcode-cn.com/problems/valid-sudoku/

題目描述:

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。

示例:

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。
 

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:

首先還是理清題意,就是每一行,每一列,每一個小正方形都不能重復出現一個字母,如下圖所示:

所以我們最直接想到就是,就是記錄它的行,列,和小正方形的值,有重復就false

思路一:

我們用一個字典,分別記錄行,列,和小正方形!

行,列我們直接可以用數字表示,小正方形如何表示呢?

這裏,我們發現一個規律,我們可以把小正方形變成用二維數組唯一標識,比如(0,0)表示左上角那個,(1,1)表示中間那個,他們和行列的關系就是(i//3,j//3)

所以任何位置我們都能找出它在哪個行,哪個列,哪個正方形裏!

時間復雜度都是常數級的

思路二:

上面我們用的空間復雜度有點多,要想辦法改進空間復雜度,

我們有點小技巧,我們只需要用一個集合就可以搞定!

比如我們把board[i][j]

用字符串:

表示行:(i) + board[i][j]

表示列: board[i][j] + (j)

表示小正方形:(i) + board[i][j] + (j)

就直接可以用一個集合搞定!

關註我的知乎專欄,了解更多的解題技巧,大家共同進步!

代碼:

思路一

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        from collections import defaultdict
        row = defaultdict(set)
        col = defaultdict(set)
        small_square = defaultdict(set)
        
        for i in range(9):
            for j in range(9):
                if board[i][j].isdigit():
                    if board[i][j] not in row[i]                     and board[i][j] not in col[j]                     and board[i][j] not in small_square[(i // 3,j // 3)]:
                        row[i].add(board[i][j]) 
                        col[j].add(board[i][j])
                        small_square[(i // 3, j // 3)].add(board[i][j])
                    else:
                        return False
        return True

思路二

python

class Solution:
    def isValidSudoku(self, board: List[List[str]]) -> bool:
        one_set = set()
        
        for i in range(9):
            for j in range(9):
                if board[i][j].isdigit():
                    row = "(" + str(i) + ")" + board[i][j]
                    col = board[i][j] + "(" + str(j) + ")"
                    small_square = "(" + str(i//3)+ ")" + board[i][j] +  "(" + str(j//3) + ")"
                    if row in one_set or col in one_set or small_square in one_set:
                        return False
                    one_set.update([row,col,small_square])
        return True

java

class Solution {
    public boolean isValidSudoku(char[][] board) {
        Set<String> one_set = new HashSet<>();
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] != '.') {
                    String row = "(" + i + ")" + board[i][j];
                    String col = board[i][j] + "(" + j + ")";
                    String small_square = "(" + i / 3 + ")" + board[i][j] + "(" + j / 3 + ")";
                    if (!one_set.add(row) || !one_set.add(col) || !one_set.add(small_square)) return false;
                }
            }
        }
        return true;
        
    }
}

[LeetCode] 36. 有效的數獨