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LeetCode 36. Valid Sudoku (Medium)

时间:2021-12-16  作者:amiya2021  

题目

Determine if a 9 x 9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

域名 row must contain the digits 1-9 without repetition.
域名 column must contain the digits 1-9 without repetition.
域名 of the nine 3 x 3 sub-boxes of the grid must contain the digits 1-9 without repetition.

Note:

  • A Sudoku board (partially filled) could be valid but is not necessarily solvable.
  • Only the filled cells need to be validated according to the mentioned rules.

 
Example 1:

Input: 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"]]
Output: true

 
Example 2:

Input: 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"]]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8\'s in the top left 3x3 sub-box, it is invalid.

 
Constraints:

  • 域名th == 9
  • board[i].length == 9
  • board[i][j] is a digit 1-9 or \'.\'.

思路

方法1 (Java)

设置一个Set集合,遍历数独表二维数组,将当前遍历到的位置的元素以3个字符串形式记录下它所在的行,列和九宫格信息,并加入Set集合。若Set集合中存在某一元素的任意行/列/九宫格信息,则认为该表不符合规则。
设当前元素值为val,位置为(i, j),位于第i/3*3+j/3个九宫格(从0开始),则该元素的信息可以设置为:
行信息:"val in row i"
列信息:"val in col j"
九宫格信息:"val in box i/3*3+j/3"

class Solution {
    public boolean isValidSudoku(char[][] board) {
        HashSet<String> set = new HashSet<>();
        for(int i = 0; i < 9; i++){
            for(int j = 0; j < 9; j++){
                if(board[i][j] == \'.\') continue;
                String row = board[i][j] + " in row " + i;
                String col = board[i][j] + " in col " + j;
                String box = board[i][j] + " in box " + (i/3*3+j/3);
                if(域名ains(row) || 域名ains(col) || 域名ains(box)) return false;
                域名(row);
                域名(col);
                域名(box);
            }
        }
        return true;
    }
}

方法2(Java)

0-8,对每一个i,设置3个boolean数组分别检测第i行、第i列和第i个九宫格,数组的index代表被检测的值。
对于每个i的值,j0-8意味着检测:
i行, 第j个数是否已位于行i中;
i列,第j个数是否已位于列i中;
i个九宫格中,该九宫格从左上角到右下角的第j数是否已位于该九宫格中。
 
设置3个数组,
boolean[] checkRow 检查位置(i, j)的值是否已存在第i行
boolean[] checkCol 检查位置(j, i)的值是否已存在第i列
boolean[] checkBox 检查位置(i/3*3+j/3, j%3*3+j%3)的值是否已存在第i个九宫格中

例如当i为4,j从0-8,对应行、列、九宫格的遍历过程

  0 1 2   3 4 5   6 7 8 
0 o o o | o 1 o | o o o
1 o o o | o 2 o | o o o 
2 o o o | o 3 o | o o o
------------------------
3 o o o | o 4 o | o o o              
4 1 2 3 | 4 5 6 | 7 8 9
5 o o o | o 6 o | o o o
------------------------
6 o o o | o 7 o | o o o
7 o o o | o 8 o | o o o
8 o o o | o 9 o | o o o

第4个九宫格中
1 2 3 
4 5 6
7 8 9
class Solution {
    public boolean isValidSudoku(char[][] board) {
        for(int i = 0; i < 9; i++){
            boolean[] checkRow = new boolean[9];
            boolean[] checkCol = new boolean[9];
            boolean[] checkBox = new boolean[9];
            for(int j = 0; j < 9; j++){
                if(board[i][j] == \'.\') {}
                else if(checkRow[board[i][j]-\'1\']) return false;
                else checkRow[board[i][j]-\'1\'] = true;
                
                if(board[j][i] == \'.\'){}
                else if(checkCol[board[j][i]-\'1\']) return false;
                else checkCol[board[j][i]-\'1\'] = true;
                
                int m = i/3*3+j/3, n = i%3*3+j%3;
                if(board[m][n] == \'.\'){}
                else if(checkBox[board[m][n]-\'1\']) return false;
                else checkBox[board[m][n]-\'1\'] = true;
            }
        }
        return true;
    }
}

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