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。
 

解题思路:

我们可以使用三个二维数组来快速判断数独是否符合要求

核心思路:使用数独中的数值作为下标,快速判断该下标位置是否被使用。

三个数组分别为:行数组(rows),列数组(coulmns),格子数组(boxes)

他们分别用于检测 每行是否有重复数字,每列是否有重复数字,每格是否有重复数字

以行数组举例:

行数组的

一号下标为行的序号(该数字位于数独的第几行)

二号下标为具体数字(0-9)

当扫描到该行的某数字时(如第一行的5),那么rows[0][5]++

如果数独在第一行有错误,出现了两个5,那么在扫描完第二个5时,rows[0][5]应该等于2,说明数独有错误,不满足要求

如果该行只有一个5或者没有5,那么rows[0][5]应该小于2,数独未出错

同理该数组应能检测整个数独的所有行的所有数值是否有重复

同理在使用coulmns时,以列序号作为第一下标,可以检测所有的列是否有数值重复

将每个3*3的盒子进行标号,也应该能检测到盒子的数值是否重复

注意点:盒子的第一下标标号

在对盒子进行标号时,我们需要每个3*3的盒子内元素通过计算能位于同一盒子中,下标相同

于是有该公式 boxIndex = i / 3 * 3 + j / 3;

可以对数独进行盒子分组
class Solution { public boolean isValidSudoku(char[][] board) { int[][] rows =
new int[9][10]; int[][] columns = new int[9][10]; int[][] boxes = new
int[9][10]; for(int i = 0;i < 9;i++){ for(int j = 0;j < 9;j++){
if(board[i][j]!='.'){ int num = board[i][j] - '0'; int boxIndex = i / 3 * 3 + j
/ 3; rows[i][num] += 1; columns[j][num] += 1; boxes[boxIndex][num] += 1;
if(rows[i][num]==2||columns[j][num]==2||boxes[boxIndex][num]==2) return false;
} } } return true; } }
时间复杂度O(n^2)

执行用时 : 34 ms, 在Valid Sudoku的Java提交中击败了68.50% 的用户

内存消耗 : 54.2 MB, 在Valid Sudoku的Java提交中击败了0.88% 的用户

 

技术
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