LeetCode No36. 有效的数独
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。
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:
![](https://assets.leetcode-cn.com/aliyun-lc-upload/uploads/2021/04/12/250px-sudoku-by-l2g-20050714svg.png)
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 == 9
board[i].length == 9
board[i][j] 是一位数字(1-9)或者 '.'
思路
暴力模拟
数独的定义是行和列都是没有重复的数,那么可以在初始的时候就直接检查行和列是否满足条件,然后再每个小矩阵的起点(小矩阵左上角),查看是否有重复的数即可。
哈希
暴力之后看了题解才想到的原来可以用哈希的思想,将每行,每列,每个小矩阵都哈希到一个数组中,如果出现有重复的数,则是不合格的。
AC代码
暴力模拟
点击查看代码
class Solution {
char[][] board;
private boolean initCheck() {
for(int i=0; i<9; i++) {
boolean[] rowOk = new boolean[10];
boolean[] colOk = new boolean[10];
for(int j=0; j<9; j++) {
if(board[i][j]!='.') {
int numR = (int) (board[i][j] - '0');
if( rowOk[numR] ) {
return false;
}
rowOk[numR] = true;
}
if( board[j][i]!='.') {
int numC = (int) (board[j][i] - '0');
if( colOk[numC] ) {
return false;
}
colOk[numC] = true;
}
}
}
return true;
}
private boolean isLittleSudo(int row, int col) {
// System.out.println("row = " + row + " col" + col);
boolean[] arr = new boolean[10];
int endR = row + 3;
int endC = col + 3;
// System.out.println("endR = " + endR + " endC" + endC);
for(int i=row; i<endR; i++) {
for(int j=col; j<endC; j++) {
// if( row==0 && col==3 ) {
// System.out.print(board[i][j]);
// }
if(board[i][j]=='.') {
continue;
}
int num = (int) (board[i][j] - '0');
if( arr[num] ) {
return false;
}
arr[num] = true;
}
// System.out.println();
}
return true;
}
public boolean isValidSudoku(char[][] board) {
this.board = board;
boolean flag = initCheck();
if( !flag ) {
return false;
}
for(int i=0; i<9; i+=3) {
for(int j=0; j<9; j+=3) {
if( !isLittleSudo(i,j) ) {
return false;
}
}
}
return true;
}
}
哈希
点击查看代码
class Solution {
public boolean isValidSudoku(char[][] board) {
boolean[][] rowOk = new boolean[10][10];
boolean[][] colOk = new boolean[10][10];
boolean[][][] littleSudo = new boolean[3][3][10];
for(int i=0; i<9; i++) {
for(int j=0; j<9; j++) {
char ch = board[i][j];
if( ch == '.' ) {
continue;
}
int num = (int)( board[i][j]-'0');
if( rowOk[i][num] || colOk[j][num] || littleSudo[i/3][j/3][num] ) {
return false;
}
rowOk[i][num] = true;
colOk[j][num] = true;
littleSudo[i/3][j/3][num] = true;
}
}
return true;
}
}