LeetCode 36. Valid Sudoku

最新推荐文章于 2024-07-04 19:49:09 发布

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本文介绍了一种验证数独是否有效的算法实现。该方法通过检查数独的每一行、每一列以及每一个九宫格内数字是否重复来判断数独是否有效。文章提供了详细的C++代码实现。

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Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

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

A partially filled sudoku which is valid.

class Solution {
private:
    bool isValidSudoku(vector<vector<char>> board, int x, int y) {
        bool flags[9] = {0};
        for(int i = 0; i < 9; ++i) {
            if(board[x][i] == '.') continue;
            if(board[x][i] >= '1' && board[x][i] <= '9') {
                if(!flags[board[x][i] - '1']) flags[board[x][i] - '1'] = true;
                else return false;
            }
        }
        memset(flags, 0, 9);
        for(int j = 0; j < 9; j++) {
            if(board[j][y] == '.') continue;
            if(board[j][y] >= '1' && board[j][y] <= '9') {
                if(!flags[board[j][y] - '1']) flags[board[j][y] - '1'] = true;
                else return false;
            }
        }
        memset(flags, 0, 9);
        int xx = x / 3 * 3;
        int yy = y / 3 * 3;
        for(int i = 0; i < 3; ++i) {
            for(int j = 0; j < 3; ++j) {
                if(board[xx + i][yy + j] == '.') continue;
                if(board[xx + i][yy + j] >= '1' && board[xx + i][yy + j] <= '9') {
                    if(!flags[board[xx + i][yy + j] - '1']) flags[board[xx + i][yy + j] - '1'] = true;
                    else return false;
                }
            }
        }
        return true;
    }
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        for(int i = 0; i < 9; ++i) {
            for(int j = 0; j < 9; ++j) {
                if(!isValidSudoku(board, i, j)) return false;
            }
        }
        return true;
    }
};