leetcode题解-36.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.
其实就是检验数独棋盘中已经给出的数字是否满足数独规则。即每行、每列、每个小九宫格内的数字是否重复。
思路：遍历棋盘，分别考察行、列、九宫格内数字是否满足条件，共九行，九列，九个九宫格。
1，使用三个hashset分别存储三个维度

    public boolean isValidSudoku1(char[][] board) {        for(int i=0; i<9; i++)        {            Set<Character> row = new HashSet<Character>();            Set<Character> col = new HashSet<Character>();            Set<Character> cube = new HashSet<Character>();            for(int j=0; j<9; j++)            {                if(board[i][j] != '.')                    if(!row.add(board[i][j]))                        return false;                if(board[j][i] != '.')                    if(!col.add(board[j][i]))                        return false;                int RowIndex = 3*(i/3);                int ColIndex = 3*(i%3);                if(board[RowIndex + j/3][ColIndex + j%3] != '.')                    if(!cube.add(board[RowIndex+j/3][ColIndex + j%3]))                        return false;            }        }        return true;    }

2，使用数组存储。这种方法速度更快。

public boolean isValidSudoku(char[][] board) {    int [] vset = new int [9];    int [] hset = new int [9];    int [] bckt = new int [9];    int idx = 0;    for (int i = 0; i < 9; i++) {        for (int j = 0; j < 9; j++) {            if (board[i][j] != '.') {                idx = 1 << (board[i][j] - '0') ;                if ((hset[i] & idx) > 0 ||                    (vset[j] & idx) > 0 ||                    (bckt[(i / 3) * 3 + j / 3] & idx) > 0) return false;                hset[i] |= idx;                vset[j] |= idx;                bckt[(i / 3) * 3 + j / 3] |= idx;            }        }    }    return true;}