Leetcode题解之数组(8)有效的数独

题目:https://leetcode-cn.com/problems/valid-sudoku/description/

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。

  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。

思路:这道题自己想没想到,想过两个for循环 暴力破解 时间复杂度O(N^2);看了下别人的思路。用二维Boolean型数组来作三个状态表:分别是行表、列表、九宫格表。int index=i/3*3+j/3  这个公式能算出 Board[i][j]所在的九宫格。

都用第一个下标作为 位置标致。

markrow[ ]  [ temp ]  从0到8表示棋盘的九行。

markrow[ ]  [ temp ] 的 i 表示 0到8列。

markblock[ index ]  [ temp ] 的i 表示0到8号九宫格。

注意:布尔数组初始化值为false.

所以该表开始都是false表示该位置为空。知道检索到temp一次之后 ,将该位置设为true。如果同一个i或j或者index 遇到过同一个temp两次表示数独有问题,将该位置的指改为true ,返回false。

代码:

class Solution { 
    public boolean isValidSudoku(char[][]board){  
        boolean markrow[][]=new boolean[9][10];  
        boolean markcow[][]=new boolean[9][10];  
        boolean markblock[][]=new boolean[9][10];
        
        for(int i=0;i<9;i++) for(int j=0;j<9;j++){ 
            int index=i/3*3+j/3;  
            if(board[j][i]!='.'){  
                int temp=board[j][i]-'0'; 
                if(markrow[j][temp])return false;  
                else markrow[j][temp]=true;   
                if(markcow[i][temp])return false;  
                else markcow[i][temp]=true;   
                if(markblock[index][temp])return false; 
                else markblock[index][temp]=true; 
            }
           }
        
        return true;   
    }   
}

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转载自blog.csdn.net/qq_34433210/article/details/83991188