LeetCode Valid Sudoku
2014-07-22 21:50
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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:
还有种更简单的解法:
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; }
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