leetcode36. Valid Sudoku

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问题描述:

Determine if a 9x9 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 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

技术分享图片
A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character ‘.‘.

Example 1:

Input:
[
  ["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:
[
  ["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.

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.
  • The given board contain only digits 1-9 and the character ‘.‘.
  • The given board size is always 9x9.

 

问题思路:

每个位置可能构成3个数独组合:行,列,块,我们就这三个进行检查。

代码:

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int usedR[9][9] = {0}, usedC[9][9] = {0}, usedB[9][9] = {0};
        for(int i = 0; i < 9; i++){
            for(int j = 0; j < 9; j++){
                if(board[i][j] != .){
                    int num = board[i][j] - 0 - 1;
                    int k = i/3 * 3 + j/3;
                    if(usedR[i][num] || usedC[j][num] || usedB[k][num])
                        return false;
                    usedR[i][num] = 1;
                    usedC[j][num] = 1;
                    usedB[k][num] = 1;
                }
            }
        }
        return true;
    }
};

 


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