LeetCode Valid Sudoku

Q:

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.

题目大意是判断输入的数字能否构成一个九宫格，即每行每列和每个方块中是否不重复地包含且只包含1-9这9个数字，这里我们使用了一个row和col的数组来模拟这个九宫格中的行和列，先把他们都初始化为0，每行每列地遍历下来如果有数字则将row相应的值+1。比如第一行第一个空格中是5这个数字，那么row[4]++，因为索引从0开始所以是row[4]++，如果这一行遍历下来还有个5 (row[4] > 0) 那么就直接return false。

代码：

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

还有种更简单的解法：

bool isValidSudoku(vector<vector<char> > &board) {        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;    }