LintCode 389: Valid Sudoku

最新推荐文章于 2020-01-13 03:26:53 发布

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本文介绍了一种算法，用于验证一个部分填充的数独是否有效。通过检查每一行、每一列以及每一个九宫格内数字是否重复，确保数独遵循基本规则。提供了两种解法，一种使用哈希表，另一种使用位标记。

Valid Sudoku

Determine whether a Sudoku is valid.

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

Example
Example 1:

Input:
[“53…7…”,“6…195…”,".98…6.",“8…6…3”,“4…8.3…1”,“7…2…6”,".6…28.","…419…5","…8…79"]
Output: true
Explanation:
The sudoku is look like this. It’s vaild.
Valid Sudoku

Example 2:

Input:
[“53…7j…”,“6…195…”,".98…6.",“8…6…3”,“4…8.3…1”,“7…2…6”,".6…28.","…419…5","…8…79"]
Output: false
Explanation:
The sudoku is look like this. It’s invaild because there are two ‘5’ in the first row and the sixth line.
image

Clarification
What is Sudoku?

http://sudoku.com.au/TheRules.aspx
https://zh.wikipedia.org/wiki/數獨
https://en.wikipedia.org/wiki/Sudoku
http://baike.baidu.com/subview/961/10842669.htm
Notice
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

解法1：
注意：
用gridIndex = (i / 3) * 3 + j / 3得到cell的index

class Solution {
public:
    /**
     * @param board: the board
     * @return: whether the Sudoku is valid
     */
    bool isValidSudoku(vector<vector<char>> &board) {
        int m = board.size();
        int n = board[0].size();
        if (m != 9 || n !=9) return false;
        
        unordered_map<char, int> um;   //number, count
        for (int i = 0; i < 9; ++i) {
            um.clear();
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] >= '0' && board[i][j] <= '9' && um[board[i][j]]++ > 0) {
                    return false;
                }
            }
        //    if (um.size() != 9) return false;        
        }

        for (int i = 0; i < 9; ++i) {
            um.clear();
            for (int j = 0; j < 9; ++j) {
                if (board[j][i] >= '0' && board[j][i] <= '9' && um[board[j][i]]++ > 0) {
                    return false;
                }
            }
       //     if (um.size() != 9) return false;        
        }

        vector<unordered_set<int>> gridArray(9); //gridIndex, number
        
        for (int i = 0; i < 9; ++i) {
            //um.clear();
            for (int j = 0; j < 9; ++j) {
                int gridIndex = (i / 3) * 3 + j / 3; // the x-th small grid

                if (board[i][j] >= '0' && board[i][j] <= '9' && gridArray[gridIndex].find(board[i][j]) != gridArray[gridIndex].end())) {
                    return false;
                } else {
                    gridArray[gridIndex].insert(board[i][j]);
                }
            }
          //  if (um.size() != 9) return false;
        }
        
        return true;
    }
};

解法2：
把解法1的三个步骤放到一起，不过要用3个2D vector。

class Solution {
public:
    /**
     * @param board: the board
     * @return: whether the Sudoku is valid
     */
    bool isValidSudoku(vector<vector<char>> &board) {
        int m = board.size();
        int n = board[0].size();
        if (m != 9 || n !=9) return false;
        vector<vector<bool>> rowsFlag(9, vector<bool>(9));
        vector<vector<bool>> colsFlag(9, vector<bool>(9));
        vector<vector<bool>> cellsFlag(9, vector<bool>(9));
        
        for (int i = 0; i < 9; ++i) {
            for (int j = 0; j < 9; ++j) {
                if (board[i][j] == '.') continue;
                int cellIndex = (i / 3) * 3 + j / 3; // the x-th small grid
                int num = board[i][j] - '1';
                if (rowsFlag[i][num] || colsFlag[j][num] || cellsFlag[cellIndex][num]) return false;
                rowsFlag[i][num] = true;
                colsFlag[j][num] = true;
                cellsFlag[cellIndex][num] = true;
            }
        }
        
        return true;
    }
};