LeetCode:Valid Sudoku

问题描述：

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.

思路：

每一行每一列都判断

代码：

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int row[9],col[9];
        for(int i = 0; i < 9; ++i){
            memset(row,0,sizeof(int) * 9);
            memset(col,0,sizeof(int) * 9);
            for(int j = 0;j < 9;++j){
                if(board[i][j] != '.'){
                    if(row[board[i][j] - '1'] > 0) return false;
                    else row[board[i][j] - '1']++;
                }
                if(board[j][i] != '.'){
                    if(col[board[j][i] - '1'] > 0) return false;
                    else col[board[j][i] - '1']++;
                }
                
            }
        }
        for(int i = 0;i < 9;i += 3){
            for(int j = 0; j < 9;j += 3){
                memset(row,0,sizeof(int) * 9);
                for(int a = 0; a < 3; ++a){
                    for(int b = 0;b < 3; ++b){
                        if(board[i + a][j + b] != '.'){
                            if(row[board[a + i][b + j] - '1'] > 0)   return false;
                            else  row[board[a + i][b + j] - '1']++;
                        }
                    }
                }
            }
        }
        return true;
    }
};

简化版本：

bool isValidSudoku(vector<vector<char> > &board) {  
        // Note: The Solution object is instantiated only once.  
        vector<vector<bool>> rows(9, vector<bool>(9,false));  
        vector<vector<bool>> cols(9, vector<bool>(9,false));  
        vector<vector<bool>> blocks(9, vector<bool>(9,false));  
  
        for(int i = 0; i < 9; i++)  
            for(int j = 0; j < 9; j++)  
            {  
                if(board[i][j] == '.')continue;  
                int num = board[i][j] - '1';  
                if(rows[i][num] || cols[j][num] || blocks[i - i%3 + j/3][num])  
                    return false;  
                rows[i][num] = cols[j][num] = blocks[i - i%3 + j/3][num] = true;  
            }  
        return true;  
    }