10、有效的数独
10、有效的数独
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。
答案解析:
import java.util.*;
class Solution {
public boolean isValidSudoku(char[][] board) {
int row = board.length;//横着的行
int col = board[0].length;//竖着的列
if (row != 9||col != 9){
return false;
}
Map<Character,Boolean> map = new HashMap<>();//键为1-9,值为true或者false
//检查row
for (int i = 0; i < row; i++) {
map.clear();
for (int j = 0; j < col; j++) {
char cur = board[i][j];
if (cur != '.'){
if (map.containsKey(cur)){
return false;
}else {
map.put(cur, true);
}
}
}
}
//检查col
for (int i = 0; i < col; i++) {
map.clear();
for (int j = 0; j < row; j++) {
char cur = board[j][i];
if (cur != '.'){
if (map.containsKey(cur)){
return false;
}else {
map.put(cur, true);
}
}
}
}
//检查9宫格
//1,2,3
//4,5,6
//7,8,9
for (int i = 0; i < 9; i++) {
int rowBegin = (i%3)*3;//行开始
int colBegin = i - rowBegin/3;//列开始
map.clear();
for (int j = rowBegin; j < rowBegin + 3; j++) {
for (int k = colBegin; k < colBegin + 3; k++) {
char cur = board[j][k];
if (cur != '.'){
if (map.containsKey(cur)){
return false;
}else {
map.put(cur, true);
}
}
}
}
}
return true;
}
}