LeetCode 36 Valid Sudoku (O(1)空间 位运算)

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.

题目分析：由于一次最多只需要检查9个数字，1 << 9 < INT_MAX，因此可以用整型数字的二进制位来判断某个数字是否出现过，击败95%

public class Solution {    public boolean isValidSudoku(char[][] board) {        int mask = 0, num = 0;        for (int i = 0; i < 9; i ++) {            mask = 0;            for (int j = 0; j < 9; j ++) {                if (board[i][j] != '.') {                    num = board[i][j] - '0';                    if ((mask & (1 << num)) == 0) {                        mask |= (1 << num);                    } else {                        return false;                    }                }            }        }        for (int j = 0; j < 9; j ++) {            mask = 0;            for (int i = 0; i < 9; i ++) {                if (board[i][j] != '.') {                    num = board[i][j] - '0';                    if ((mask & (1 << num)) == 0) {                        mask |= (1 << num);                    } else {                        return false;                    }                }            }        }        for (int i = 0; i < 3; i ++) {            for (int j = 0; j < 3; j ++) {                int x = i * 3;                int y = j * 3;                mask = 0;                for (int k = 0; k < 3; k ++) {                    for (int l = 0; l < 3; l ++) {                        if (board[x + k][y + l] != '.') {                            num = board[x + k][y + l] - '0';                            if ((mask & (1 << num)) == 0) {                                mask |= (1 << num);                            } else {                                return false;                            }                        }                    }                }            }        }        return true;    }}