LeetCode 36. Valid Sudoku

历史记录

清除记录

猜你想搜

AcWing热点
App
登录/注册

LeetCode 36. Valid Sudoku 原题链接中等

作者：

wzc1995 , 2018-04-18 16:00:15 , 所有人可见 , 阅读 2945

5

7

题目描述

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

样例

输入:
[
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
输出: true

输入:
[
  ["8","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
输出: false
解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与示例 1 相同。
     但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

算法

(位运算判重)

分别使用一个整型数组记录每行、每列和每个九宫格内数字的存在情况。
位运算可以极大的简化判断，提高效率，具体看代码。

C++ 代码

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        vector<int> row(9), col(9), squ(9); // 使用三个整型数组判重。
        for (int i = 0; i < 9; i++)
            for (int j = 0; j < 9; j++) {
                if (board[i][j] == '.')
                    continue;
                if (board[i][j] < '1' || board[i][j] > '9') return false;
                int num = board[i][j] - '0';

                // 以row[i] & (1 << num)为例，这是判断第i行中，num数字是否出现过。
                // 即row[i]值的二进制表示中，第num位是否是1。
                // 以下col和squ同理。

                if ((row[i] & (1 << num)) ||
                    (col[j] & (1 << num)) ||
                    (squ[(i / 3) * 3 + (j / 3)] & (1 << num)))
                    return false;

                row[i] |= (1 << num);
                col[j] |= (1 << num);
                squ[(i / 3) * 3 + (j / 3)] |= (1 << num);
            }
        return true;
    }
};

7 评论

迷弟 2021-03-02 18:01

看不太懂。。可读性不太行呀(远古时期炫技石锤hh😁)

该用户被封禁 2021-03-03 00:01

这个位运算帅呆了 lol

dys 2020-06-15 12:12

位运算那一块看的好懵

Microlin 2018-07-22 22:46

下面是使用set的方法； map解法类似

class Soulation
{
public:
    boolbool isValidSudoku(vector<vector<char>>& board) {
        for (int i = 0; i < 9; i++) {
            set<char> row;
            set<char> column;
            set<char> cube;
            for (int j = 0; j < 9; j++) {
                // 检查第i行，在横坐标位置
                if (board[i][j] != '.' && !*row.insert(board[i][j]).second)
                    return false;
                // 检查第i列，在纵坐标位置
                if (board[j][i] != '.' && !*column.insert(board[j][i]).second)
                    return false;
                // 行号+偏移量
                int RowIndex = 3 * (i / 3) + j / 3;
                // 列号+偏移量
                int ColIndex = 3 * (i % 3) + j % 3;
                //每个小九宫格，总共9个
                if (board[RowIndex][ColIndex] != '.'
                    && !*cube.insert(board[RowIndex][ColIndex]).second)
                    return false;
            }
        }
        return true;
    }
};