题目:
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 ‘.‘.
![这里写图片描述](http://img.blog.csdn.net/20160409183641502)
A partially filled sudoku which is valid.
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
思路:
- 首先这是一道数独题目,关于数独的性质,可以参考下面的链接
- 考虑根据数独的各个条件来逐个解决,需要判断每行,每一列,以及每个小方格是否是1~9,出现一次,思路是建立一个list添加,判断是否出现重复,出现返回false;同样如果board == null,或者行数和列数不等于9,也返回false
-
代码:
public class Solution {
public boolean isValidSudoku(char[][] board) {
ArrayList<ArrayList<Character>> rows = new ArrayList<ArrayList<Character>>();
ArrayList<ArrayList<Character>> cols = new ArrayList<ArrayList<Character>>();
ArrayList<ArrayList<Character>> boxs = new ArrayList<ArrayList<Character>>();
if(board == null || board.length != 9|| board[0].length != 9){
return false;
}
for(int i = 0;i < 9;i++){
rows.add(new ArrayList<Character>());
cols.add(new ArrayList<Character>());
boxs.add(new ArrayList<Character>());
}
for(int a = 0;a < board.length;a++){
for(int b = 0;b < board[0].length;b++){
if(board[a][b] == ‘.‘){
continue;
}
ArrayList<Character> row = rows.get(a);
if(row.contains(board[a][b])){
return false;
}else{
row.add(board[a][b]);
}
ArrayList<Character> col = cols.get(b);
if(col.contains(board[a][b])){
return false;
}else{
col.add(board[a][b]);
}
ArrayList<Character> box = boxs.get(getNum(a,b));
if(box.contains(board[a][b])){
return false;
}else{
box.add(board[a][b]);
}
}
}return true;
}
public int getNum(int i,int j){
return (i/3)*3+j/3;
}
}
时间: 2024-11-25 13:29:33