【Leetcode】036. Valid Sudoku

题目：

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.

Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

题解：

Solution 1

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        const int n = 9;

        for(int i = 0; i < n; ++i){
            bool used[n] = {false};
            for(int j = 0; j < n; ++j){
                if(isused(board[i][j], used))
                    return false;
            }
        }
        for(int i = 0; i < n; ++i){
            bool used[n] = {false};
            for(int j = 0; j < n; ++j){
                if(isused(board[j][i], used))
                    return false;
            }
        }
        for(int r = 0; r < 3; ++r){
            for(int c = 0; c < 3; ++c){
                bool used[n] = {false};
                for(int i = r * 3; i < r * 3 + 3; ++i){
                    for(int j = c * 3; j < c * 3 + 3; ++j){
                        if(isused(board[i][j], used))
                           return false;
                    }
                }
            }
        }
        return true;
    }
    bool isused(char val, bool used[9]){
        if(val == ‘.‘)
            return false;
        if(used[val - ‘1‘])
            return true;
        used[val - ‘1‘] = true;
        return false;
    }
};    

Solution 2

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        const int n = 9;
        vector<vector<bool>> rowboard(n, vector<bool>(n, false));
        vector<vector<bool>> colboard(n, vector<bool>(n, false));
        vector<vector<bool>> celboard(n, vector<bool>(n, false));
        for(int i = 0; i < n; ++i){
            for(int j = 0; j < n; ++j){
                if(board[i][j] == ‘.‘) continue;
                int ch = board[i][j] - ‘1‘;
                if(rowboard[i][ch] || colboard[ch][j] || celboard[3 * (i / 3) + j / 3][ch])
                    return false;
                rowboard[i][ch] = true;
                colboard[ch][j] = true;
                celboard[3 * (i / 3) + j / 3][ch] = true;
            }
        }
        return true;
    }
};