知识拓展练习(经典面试题) / 34. 有效的数独
一、题目
请你判断一个 9 x 9 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。(请参考示例图)
注意:
①一个有效的数独(部分已被填充)不一定是可解的。
②只需要根据以上规则,验证已经填入的数字是否有效即可。
③空白格用 '.' 表示。
二、示例
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。
三、提示
board.length == 9
board[i].length == 9
board[i][j] 是一位数字(1-9)或者 '.'
四、参考题解
1、方法一:一次遍历
有效的数独满足以下三个条件:
①同一个数字在每一行只能出现一次;
②同一个数字在每一列只能出现一次;
③同一个数字在每一个小九宫格只能出现一次。
可以使用哈希表记录每一行、每一列和每一个小九宫格中,每个数字出现的次数。只需要遍历数独一次,在遍历的过程中更新哈希表中的计数,并判断是否满足有效的数独的条件即可。
对于数独的第 i 行第 j 列的单元格,其中 0≤i,j<9,该单元格所在的行下标和列下标分别为 i 和 j,该单元格所在的小九宫格的行数和列数分别为 ⌊i/3⌋ 和 ⌊j/3⌋,其中 0≤⌊i/3⌋,⌊j/3⌋<3。
由于数独中的数字范围是 1 到 9,因此可以使用数组代替哈希表进行计数。
具体做法是,创建二维数组 rows 和 columns 分别记录数独的每一行和每一列中的每个数字的出现次数,创建三维数组 subboxes 记录数独的每一个小九宫格中的每个数字的出现次数,其中 rows[i][index]、columns[j][index] 和 subboxes[⌊
i/3][⌊j/3⌋][index] 分别表示数独的第 i 行第 j 列的单元格所在的行、列和小九宫格中,数字 index+1 出现的次数,其中 0≤index<9,对应的数字 index+1 满足 1≤index+1≤9。
如果 board[i][j] 填入了数字 n,则将 rows[i][n−1]、columns[j][n−1] 和 subboxes[⌊i3⌋][⌊j3⌋][n−1] 各加 1。如果更新后的计数大于 1,则不符合有效的数独的条件,返回 false。
如果遍历结束之后没有出现计数大于 1 的情况,则符合有效的数独的条件,返回 true。
//Java
class Solution {
public boolean isValidSudoku(char[][] board) {
int[][] rows = new int[9][9];
int[][] columns = new int[9][9];
int[][][] subboxes = new int[3][3][9];
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char c = board[i][j];
if (c != '.') {
int index = c - '0' - 1;
rows[i][index]++;
columns[j][index]++;
subboxes[i / 3][j / 3][index]++;
if (rows[i][index] > 1 || columns[j][index] > 1 || subboxes[i / 3][j / 3][index] > 1) {
return false;
}
}
}
}
return true;
}
}
复杂度分析
时间复杂度:O(1)。数独共有 81 个单元格,只需要对每个单元格遍历一次即可。
空间复杂度:O(1)。由于数独的大小固定,因此哈希表的空间也是固定的。