leetcode 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。

  1. 数字 1-9 在每一行只能出现一次。
  2. 数字 1-9 在每一列只能出现一次。
  3. 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。

暴力求解:

class Solution {
public:
    bool isValidSudoku(vector<vector<char>>& board) {
        set<int> s1;
        int i,j,k,m;
        for(i=0;i<9;i++){
            for(int j=0;j<9;j++){
                if(board[i][j]!='.'){
                    if(s1.find(board[i][j])!=s1.end())
                        return false;
                    else s1.insert(board[i][j]);
                }
            }
            s1.clear(); 
        }
        for(j=0;j<9;j++){
            for(int i=0;i<9;i++){
                if(board[i][j]!='.'){
                    if(s1.find(board[i][j])!=s1.end())
                        return false;
                    else s1.insert(board[i][j]);
                }
            }
            s1.clear(); 
        }
        for(i=0;i<9;i+=3){
            for(j=0;j<9;j+=3){
                for(k=i;k<i+3;k++){
                    for(m=j;m<j+3;m++){
                        if(board[k][m]!='.'){
                            if(s1.find(board[k][m])!=s1.end())
                                return false;
                            else s1.insert(board[k][m]);
                        }
                    }
                }
                s1.clear();
                
            }
        }
        return true;
        
    }
    
};

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转载自blog.csdn.net/weixin_41582705/article/details/81357393