有效的数独 LeetCode(数组的巧妙运用)

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本文介绍了一种有效的数独验证算法，通过使用map和三维数组优化检查过程，确保数独中1-9数字在每行、每列及每个3x3宫内不重复。文章对比了两种实现方式，并分析了其运行效率。

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有效的数独

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。

示例 1:

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

示例 2:

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。

说明:

一个有效的数独（部分已被填充）不一定是可解的。
只需要根据以上规则，验证已经填入的数字是否有效即可。
给定数独序列只包含数字 1-9 和字符 '.' 。
给定数独永远是 9x9 形式的。

题解:

今天刷了好几道题，就这题还有点意思，主要是优化，横纵坐标的有效性没问题直接用map搞，3x3的方格的有效性我一开始是直接写了个check函数，在函数里用map进行验证，后来看见别人的代码，觉得很巧妙，可以开一个三维数组，假设验证第i行第j个，那么可以把它唯一标记为p[i/3][j/3]，可以验证(0,0),(0,1)一直到(2,2)也就是第一个方格所有坐标都符合这个唯一标记，而其他的坐标不符合，那么开一个三维数组就可以标记整张图的每个3x3的方格里的有效性。我修改之后的时间居然跑得和原来的一样。。莫名其妙

我修改前的代码:

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        for(int i=0;i<9;i++)
        {
            map<char,int>p;
            map<char,int>p1;
            for(int j=0;j<9;j++)
            {
                if(i%3==0&&j%3==0)
                {
                    if(!check(board,i,j))
                        return false;
                }
                if(board[i][j]!='.')
                {
                    if(p[board[i][j]])
                        return false;
                    p[board[i][j]]=1;
                }
                if(board[j][i]!='.')
                {
                    if(p1[board[j][i]])
                        return false;
                    p1[board[j][i]]=1;
                }
            }
        }
        return true;
    }
    bool check(vector<vector<char>>& board,int sx,int sy)
    {
        map<char,int>p;
        for(int i=sx;i<sx+3;i++)
        {
            for(int j=sy;j<sy+3;j++)
            {
                if(board[i][j]=='.')
                    continue;
                if(p[board[i][j]])
                    return false;
                p[board[i][j]]=1;
            }
        }
        return true;
    }
};

我修改后的代码:

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        int p2[10][10][10];
        memset(p2,0,sizeof(p2));
        for(int i=0;i<9;i++)
        {
            map<char,int>p;
            map<char,int>p1;
            for(int j=0;j<9;j++)
            {
                if(board[i][j]!='.')
                {
                    if(p[board[i][j]])
                        return false;
                    p[board[i][j]]=1;
                    int t=board[i][j]-'0';
                    if(p2[i/3][j/3][t])
                        return false;
                    p2[i/3][j/3][t]=1;
                }
                if(board[j][i]!='.')
                {
                    if(p1[board[j][i]])
                        return false;
                    p1[board[j][i]]=1;
                }
            }
        }
        return true;
    }
};