36. 有效的数独

请你判断一个 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。(请参考示例图)
    数独部分空格内已填入了数字,空白格用 ‘.’ 表示。

注意:

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。

示例 1:


输入:board = 
[["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:board = 
[["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。

提示:

  • board.length == 9
  • board[i].length == 9
  • board[i][j] 是一位数字或者 ‘.’

解题思路

  1. 按行遍历
  2. 按列遍历
  3. 按九宫格遍历
    保证每一次遍历都不会产生重复的数字,即可视为有效的数独

代码

class Solution {
public boolean isValidSudoku(char[][] board) {
        int n=board.length,m=board[0].length;
        for(int i=0;i<n;i++)
        {
            Set<Character> set=new HashSet<>();
            for(int j=0;j<m;j++)
                if(board[i][j]!='.')
                if(set.contains(board[i][j]))
                    return false;
                else set.add(board[i][j]);
        }
        for(int j=0;j<m;j++)
        {
            Set<Character> set=new HashSet<>();
            for(int i=0;i<n;i++)
             if(board[i][j]!='.')
                if(set.contains(board[i][j]))
                    return false;
                else set.add(board[i][j]);
        }
        for(int i=0;i<n;i+=3)
            for(int j=0;j<m;j+=3)
            {
                Set<Character> set=new HashSet<>();
                for(int k=i;k<i+3;k++)
                    for(int l=j;l<j+3;l++)
                     if(board[k][l]!='.')
                 if(set.contains(board[k][l]))
                    return false;
                else set.add(board[k][l]);

            }
            return true;

    }
}
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