【Leetcode】Valid Sudoku

题目链接：https://leetcode.com/problems/valid-sudoku/

题目：

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.

思路：

每行、每列、每个黑粗线内小方块的数分别用HashSet保存判断是否唯一。

算法：

 </pre><pre name="code" class="java">    public boolean isValidSudoku(char[][] board) {
         Set<Character> row = new HashSet<Character>();
         Set<Character> column = new HashSet<Character>();
         Set<Character> subBox = new HashSet<Character>();

         for (int i = 0; i < board.length; i++) {
             for (int j = 0; j < board[0].length; j++) {
                 // System.out.print(board[i][j]+" ");
                 if (board[i][j] != '.') {
                     if (column.contains(board[i][j])) {
                         return false;
                     } else {
                         column.add(board[i][j]);
                     }
                 }
                 // System.out.println();
             }
             column.clear();
         }

         // System.out.println("=============");
         for (int i = 0; i < board.length; i++) {
             for (int j = 0; j < board[0].length; j++) {
                 if (board[j][i] != '.') {
                     if (row.contains(board[j][i])) {
                         return false;
                     } else {
                         row.add(board[j][i]);
                     }
                 }
             }
             row.clear();
         }

         for (int i = 0; i < board.length; i += 3) {
             for (int j = 0; j < board[0].length; j += 3) {
                 for (int ii = i; ii < i + 3; ii++) {
                     for (int jj = j; jj < j + 3; jj++) {
                         if (board[ii][jj] != '.') {
                             if (subBox.contains(board[ii][jj])) {
                                 return false;
                             } else {
                                 subBox.add(board[ii][jj]);
                             }
                         }
                     }
                 }
                 subBox.clear();
             }
         }
         return true;
     }