36. Valid Sudoku(合法数独)
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.
关于数独的简介:
1.Each row must have the numbers 1-9 occuring just once.
2.Each column must have the numbers 1-9 occuring just once.
3.And the numbers 1-9 must occur just once in each of the 9 sub-boxes of the grid.
题目大意:
判断一个给定的二维数组是否是一个合法的数独矩阵。
思路:
采用set这一容器,来进行去重。
1.判断每一行是否合法。
2.判断每一列是否合法。
3.判断每一个九宫格是否合法。
代码如下:
class Solution { public: bool isValidSudoku(vector<vector<char>>& board) { set<char> mySet; //1.判断每一行是否合法 for (int row = 0; row < 9; row++) { //cout<<"检测行:"<<row<<endl; for (int column = 0; column < 9; column++) { if (board[row][column] == '.') { continue; } if (mySet.find(board[row][column]) == mySet.end()) { mySet.insert(board[row][column]); } else { return false; } } mySet.clear(); } //2.判断每一列是否合法 for (int row = 0; row < 9; row++) { //cout<<"检测列:"<<row<<endl; for (int column = 0; column < 9; column++) { if (board[column][row] == '.') { continue; } if (mySet.find(board[column][row]) == mySet.end()) { mySet.insert(board[column][row]); } else { return false; } } mySet.clear(); } //3.判断每一个九宫格是否合法 for (int row = 0; row < 9; row += 3) { for (int column = 0; column < 9; column += 3) { for (int i = row; i < row + 3; i++) { for (int j = column; j < column + 3; j++) { if (board[i][j] == '.') { continue; } if (mySet.find(board[i][j]) == mySet.end()) { mySet.insert(board[i][j]); } else { return false; } } } mySet.clear(); } } return true; } };
2016-08-13 12:21:54
免责声明:本站发布的内容(图片、视频和文字)以原创、转载和分享为主,文章观点不代表本网站立场,如果涉及侵权请联系站长邮箱:is@yisu.com进行举报,并提供相关证据,一经查实,将立刻删除涉嫌侵权内容。