哈希表//有效的数独
判断一个 9x9 的数独是否有效。只需要根据以下规则,验证已经填入的数字是否有效即可。
- 数字
1-9
在每一行只能出现一次。 - 数字
1-9
在每一列只能出现一次。 - 数字
1-9
在每一个以粗实线分隔的3x3
宫内只能出现一次。
上图是一个部分填充的有效的数独。
数独部分空格内已填入了数字,空白格用 '.'
表示。
示例 1:
输入: [ ["5","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 输出: true
示例 2:
输入: [ ["8","3",".",".","7",".",".",".","."], ["6",".",".","1","9","5",".",".","."], [".","9","8",".",".",".",".","6","."], ["8",".",".",".","6",".",".",".","3"], ["4",".",".","8",".","3",".",".","1"], ["7",".",".",".","2",".",".",".","6"], [".","6",".",".",".",".","2","8","."], [".",".",".","4","1","9",".",".","5"], [".",".",".",".","8",".",".","7","9"] ] 输出: false 解释: 除了第一行的第一个数字从 5 改为 8 以外,空格内其他数字均与 示例1 相同。 但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。
说明:
- 一个有效的数独(部分已被填充)不一定是可解的。
- 只需要根据以上规则,验证已经填入的数字是否有效即可。
- 给定数独序列只包含数字
1-9
和字符'.'
。 - 给定数独永远是
9x9
形式的。
class Solution {
public boolean isValidSudoku(char[][] board) {
for(int i = 0; i < 9; i++){
HashSet<Character> line = new HashSet<>();
HashSet<Character> col = new HashSet<>();
HashSet<Character> cube = new HashSet<>();
for(int j = 0; j < 9; j++){
if(board[i][j] != '.'&&!line.add(board[i][j]))
return false;
if(board[j][i] != '.'&&!col.add(board[j][i]))
return false;
int m = i/3*3+j/3;
int n = i%3*3+j%3;
if(board[m][n] != '.'&& !cube.add(board[m][n]))
return false;
}
}
return true;
}
}
class Solution {
public boolean isValidSudoku(char[][] board) {
for(int i = 0; i < 9; i++){
int[] bit_map_row = new int[9];
int[] bit_map_col = new int[9];
int[] bit_map_cube = new int[9];
for(int j = 0; j < 9; j++){
if(board[i][j] != '.')
if(bit_map_row[board[i][j]-'1'] == 1)
return false;
else
bit_map_row[board[i][j]-'1'] = 1;
if(board[j][i] != '.')
if(bit_map_col[board[j][i]-'1'] == 1)
return false;
else
bit_map_col[board[j][i]-'1'] = 1;
int m = 3*(i/3)+j/3;
int n = 3*(i%3)+j%3;
int val = board[m][n];
if(val != '.')
if(bit_map_cube[val-'1'] == 1)
return false;
else
bit_map_cube[val-'1'] = 1;
}
}
return true;
}
}