36. Valid Sudoku 终极之牛逼方法
Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
- Each row must contain the digits
1-9
without repetition. - Each column must contain the digits
1-9
without repetition. - Each of the 9
3x3
sub-boxes of the grid must contain the digits1-9
without repetition.
The Sudoku board could be partially filled, where empty cells are filled with the character '.'
.
Example 1:
Input: [ ["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"] ] Output: true
Example 2:
Input: [ ["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"] ] Output: false Explanation: Same as Example 1, except with the 5 in the top left corner being modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.
Note:
- A Sudoku board (partially filled) could be valid but is not necessarily solvable.
- Only the filled cells need to be validated according to the mentioned rules.
- The given board contain only digits
1-9
and the character'.'
. - The given board size is always
9x9
.
终极之无敌巧妙方法,理解不能
class Solution {
/*
int rowIndex = 3 * (i / 3);
int colIndex = 3 * (i % 3);
i~[0,8]
0 1 2 3 4 5 6 7 8
0 0 0 3 3 3 6 6 6 row Index
0 3 6 0 3 6 0 3 6 colIndex
0 0 0 1 1 1 2 2 2 j / 3
0 1 2 0 1 2 0 1 2 j % 3
扫描cube时是从左到右从上到下扫描9个cube
*/ public boolean isValidSudoku(char[][] board) { for(int i = 0; i < board.length; i++){ HashSet<Character> rows = new HashSet<>(); HashSet<Character> cols = new HashSet<>(); HashSet<Character> cube = new HashSet<>(); for(int j = 0; j < board[0].length; j++){ if(board[i][j] != '.' && !rows.add(board[i][j])) return false; if(board[j][i] != '.' && !cols.add(board[j][i])) return false; int rowIndex = 3 * (i / 3); int colIndex = 3 * (i % 3); if(board[rowIndex + j / 3][colIndex + j % 3] != '.' && !cube.add(board[rowIndex + j / 3][colIndex + j % 3])) return false; } } return true; } }
好的 终于有一种看得懂的方法了,那就是遍历三次,检查三次
class Solution { public boolean isValidSudoku(char[][] board) { for(int i = 0; i < 9; i++) { if(!isvalidrow(board, i)) return false; if(!isvalidcol(board, i)) return false; for(int j = 0; j < 9; j++) { if(i % 3 == 0 && j % 3 == 0) { if(!isvalids(board, i, j)) return false; } } } return true; } public boolean isvalidrow(char[][] ch, int r) { Set<Character> set = new HashSet(); for(int i = 0; i < 9; i++) { if(ch[r][i] != '.') { if(!set.contains(ch[r][i])) set.add(ch[r][i]); else return false; } } return true; } public boolean isvalidcol(char[][] ch, int c) { Set<Character> set = new HashSet(); for(int i = 0; i < 9; i++) { if(ch[i][c] != '.') { if(!set.contains(ch[i][c])) set.add(ch[i][c]); else return false; } } return true; } public boolean isvalids(char[][] ch, int r, int c) { Set<Character> set = new HashSet(); for(int i = r; i < r + 3; i++) { for(int j = c; j < c + 3; j++) { if(ch[i][j] != '.') { if(!set.contains(ch[i][j])) set.add(ch[i][j]); else return false; } } } return true; } }
可不可以只遍历一次呢?可以,牛逼
class Solution { public boolean isValidSudoku(char[][] board) { Set seen = new HashSet(); for (int i=0; i<9; ++i) { for (int j=0; j<9; ++j) { char number = board[i][j]; if (number != '.') if (!seen.add(number + " in row " + i) || !seen.add(number + " in column " + j) || !seen.add(number + " in block " + i/3 + "-" + j/3)) return false; } } return true; } }