LeetCode 36 Valid Sudoku (O(1)空间位运算)

最新推荐文章于 2021-06-13 19:59:01 发布

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最新推荐文章于 2021-06-13 19:59:01 发布

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分类专栏： Leetcode Medium 套路思维题 LeetCode 文章标签： leetcode 位运算

本文链接：https://blog.youkuaiyun.com/Tc_To_Top/article/details/74682442

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本文介绍了一种使用二进制位标记法验证数独有效性的高效算法。该方法通过检查行、列及每个宫格内的数字是否重复来判断数独是否有效。此算法简单且快速，能够轻松应对部分填充的数独板。

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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;
    }
}