【LeetCode】36. Valid Sudoku 解題報告

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36. Valid Sudoku / Easy

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

  1. Each row must contain the digits 1-9 without repetition.
  2. Each column must contain the digits 1-9 without repetition.
  3. Each 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:

  • board.length == 9
  • board[i].length == 9
  • board[i][j] is a digit or ‘.’.

Solution

思路

要檢查數獨是否 valid，需要檢查每一行 row 、 column 還有九宮格。
最簡單的寫法就是每一行都暴力判斷，或是用 set 把每一行的數字存起來，判斷是否重複。
但我們可以用比較省空間的方式來存，數獨的長寬固定是 9，我們可以用 9 個 bits 來儲存數字是否出現的狀態。這樣每一行列僅需要 2^9 = 512 大小的空間即可。

效能

Complexity

  • Time Complexity: O(N^2)
  • Space Complexity: O(N^2)

LeetCode Result

Code

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class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        const int N = 9;
        bitset<N> rows[N];
        bitset<N> cols[N];
        bitset<N> boxes[N];
        
        for(int i = 0; i < N; ++i) {
            for(int j = 0; j < N; ++j) {
                char val = board[i][j];
                if(val == '.') continue;
                bitset<N> pos = 0;
                pos[val-'1'] = 1;
                
                if((rows[i] & pos).any()) {
                    return false;
                }
                rows[i] |= pos;
                
                if((cols[j] & pos).any()) {
                    return false;
                }
                cols[j] |= pos;
                
                int box_idx = ((i/3)*3) + (j/3);
                if((boxes[box_idx] & pos).any()) {
                    return false;
                }
                boxes[box_idx] |= pos;
            }
        }        
        return true;
    }
};