Leetcode36 Valid Sudoku

Valid Sudoku

Determine if a Sudoku is valid, according to: Sudoku Puzzles - The Rules.

The Sudoku board could be partially filled, where empty cells are filled with the character ‘.’.



Note:

A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

Solution1

最简单的方法就是用一个HashSet来分别记录每一行，每一列，每一个九小格方块。如下：

import java.util.HashSet;
public class Solution {
public boolean isValidSudoku(char[][] board) {
if(board.length!=9) return false;
if(board[0].length!=9) return false;
HashSet<Character> record = new HashSet<Character>();
for(int i=0;i<9;i++){
for(int j=0;j<9;j++){
char c = board[i][j];
if(c=='.') continue;
if(c<'0'||c>'9'||record.contains(c)) return false;
record.add(c);
}
record.clear();
}

for(int j=0;j<9;j++){
for(int i=0;i<9;i++){
char c = board[i][j];
if(c=='.') continue;
if(c<'0'||c>'9'||record.contains(c)) return false;
record.add(c);
}
record.clear();
}

for(int m=0;m<3;m++){
for(int n=0;n<3;n++){
for(int i=0;i<3;i++){
for(int j=0;j<3;j++){
char c = board[m*3+i][n*3+j];
if(c=='.') continue;
if(c<'0'||c>'9'||record.contains(c)) return false;
record.add(c);
}
}
record.clear();
}
}
return true;
}
}

这种方法简单易懂，但是要对整个数组遍历三次，这一般都是不太合适的。所以提出改进方法， 两种思路：一是改进算法，另一种一般用空间换时间。

Solution2

提出一种用空间换时间的方法，如下：

public class Solution {
public boolean isValidSudoku(char[][] board) {
if(board.length!=9) return false;
if(board[0].length!=9) return false;
boolean rows[][] = new boolean[9][9];
boolean cols[][] = new boolean[9][9];
boolean grids[][] = new boolean[9][9];
for(int i=0;i<9;i++){
for(int j=0;j<9;j++){
char c = board[i][j];
if(c=='.') continue;
if(c<'1'||c>'9'||rows[i][c-'1']||cols[j][c-'1']||grids[i/3*3+j/3][c-'1']) return false;
rows[i][c-'1'] = true;
cols[j][c-'1'] = true;
grids[i/3*3+j/3][c-'1'] = true;//这是将小方块映射到1——9的某个数
}
}
return true;
}
}