36. 有效的数独
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。
- 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)
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。
- 空白格用
'.'表示。
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 == 9board[i].length == 9board[i][j]是一位数字(1-9)或者'.'
自己的暴力解法(未解决3×3块内问题):
1 class Solution {
2 public boolean isValidSudoku(char[][] board) {
3 //遍历所有数字
4 for (int i = 0;i < 9;i++){
5 for(int j = 0;j < 9;j++){
6 int temp = board[i][j];
7 //判断行是否有重复的数字
8 for(int a = 0;a < 9;a++){
9 if(a == j){
10 continue;
11 }else if(board[i][a] == board[i][j]){
12 return false;
13 }
14 }
15 //判断列是否有重复的数字
16 for(int b = 0;b < 9;b++){
17 if(b == i){
18 continue;
19 }else if(board[b][j] == board[i][j]){
20 return false;
21 }
22 }
23 //判断3×3宫内是否有重复数字
24
25 }
26 }
27 return true;
28 }
29 }
查看题解:
下面这个图是每一个3 * 3格子所在的blockIndex
0 1 2
3 4 5
6 7 8
先看列,是按照加一的方式增加,所以是j / 3
再看行,是按照0*3 1*3 2*3 的方式增加 所以先除3算出前面的数,也就是0 1 2 ,再乘3
行 + 列 就是方格中元素的blockIndex
1 class Solution {
2 public boolean isValidSudoku(char[][] board) {
3 // 记录某行,某位数字是否已经被摆放
4 boolean[][] row = new boolean[9][9];
5 // 记录某列,某位数字是否已经被摆放
6 boolean[][] col = new boolean[9][9];
7 // 记录某 3x3 宫格内,某位数字是否已经被摆放
8 boolean[][] block = new boolean[9][9];
9
10 for (int i = 0; i < 9; i++) {
11 for (int j = 0; j < 9; j++) {
12 if (board[i][j] != '.') {
// 将字符转化为int型的整数,方便后续使用其下标。-‘1’的妙用;
13 int num = board[i][j] - '1';
// 判断数字在哪一块中,也可是i / 3 + j / 3 * 3;
14 int blockIndex = i / 3 * 3 + j / 3;
15 if (row[i][num] || col[j][num] || block[blockIndex][num]) {
16 return false;
17 } else {
18 row[i][num] = true;
19 col[j][num] = true;
20 block[blockIndex][num] = true;
21 }
22 }
23 }
24 }
25 return true;
26 }
27 }
原创文章,作者:ItWorker,如若转载,请注明出处:https://blog.ytso.com/tech/pnotes/268073.html