Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:
1-9
without repetition.1-9
without repetition.3x3
sub-boxes of the grid must contain the digits 1-9
without repetition.
A partially filled sudoku which is valid.
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:
1-9
and the character ‘.‘
.9x9
.题目大意:验证一个数独是否是有效的:某一行/一列/3x3的子格子数字不能重复。
1 class Solution { 2 public: 3 bool isValidSudoku(vector<vector<char>>& board) { 4 //map<int, map<char, int> > row, col, grid; 5 vector<vector<int> > row(9, vector<int>(10, 0)), col(9, vector<int>(10, 0)), grid(9, vector<int>(10, 0)); 6 for (int i = 0; i < 9; i++) { 7 for (int j = 0; j < 9; j++) { 8 char tmp = board[i][j]; 9 if (tmp != ‘.‘) { 10 int num = tmp - ‘0‘; 11 if (row[i][num] != 0) { 12 return false; 13 } else { 14 row[i][num] = 1; 15 } 16 17 if (col[j][num] != 0) { 18 return false; 19 } else { 20 col[j][num] = 1; 21 } 22 // 从左往右,从上至下,3x3格子分别从0-8编号。 23 if (grid[(i / 3) * 3 + (j / 3)][num] != 0) { 24 return false; 25 } else { 26 grid[(i / 3) * 3 + (j / 3)][num] = 1; 27 } 28 } 29 30 } 31 } 32 return true; 33 } 34 };
原文:https://www.cnblogs.com/qinduanyinghua/p/11521838.html