Valid Sudoku leetcode

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.

首先按照每行chek 然后每列check 最后每个block 分别check

check的方法是维护一个hashset 如果是'.'就跳过 如果是char就看set中是否包含 如果包含就return false 不包含就把当前char放入set中

对于block的i j取法比较trick 

k/ 3 * 3  ~ k / 3 * 3 + 2 是代表行数 k % 3 * 3 ~ k % 3 *3 + 2 是列数 这样对于每个k都可以表示一个block
时间复杂度 O(3*n^2)

public class Solution {
    public boolean isValidSudoku(char[][] board) {
        HashSet<Character> set = new HashSet<Character>();
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.') {
                    continue;
                }
                if (set.contains(board[i][j])) {
                    return false;
                }
                set.add(board[i][j]);
            }
            set.clear();
        }
        for (int j = 0; j < 9; j++) {
            for (int i = 0; i < 9; i++) {
               if (board[i][j] == '.') {
                    continue;
                }
                if (set.contains(board[i][j])) {
                    return false;
                }
                set.add(board[i][j]);
            }
            set.clear(); 
        }
        for (int k = 0; k < 9; k++) {
            for (int i = k / 3 * 3; i < k / 3 * 3 + 3; i ++) {
                for (int j = (k % 3) * 3; j < (k % 3) * 3 + 3; j++) {
                    if (board[i][j] == '.') {
                    continue;
                }
                if (set.contains(board[i][j])) {
                    return false;
                }
                set.add(board[i][j]);
                }
            }
            set.clear(); 
        }
        return true;
    }
}