Valid Sudoku
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Question Solution
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 '.'.
A partially filled sudoku which is valid.
分析：每行中每个元素可以为空，但0~9数字只出现只能出现一次；
每列中每个元素可以为空，但0~9数字只出现只能出现一次；
每个小九宫格中每个元素可以为空，但0~9数字只出现只能出现一次；
public class Solution {
boolean isV(int xs, int xe, int ys, int ye, char[][] board)
{
Map<Character,Integer> z=new HashMap<Character,Integer>();
for(int i=xs;i<=xe;i++)
for(int j=ys;j<=ye;j++)
if(board[i][j]>='0'&&board[i][j]<='9')
if(z.get(board[i][j])!=null)
return false;
else
z.put(board[i][j],1);
return true;
}
public boolean isValidSudoku(char[][] board) {
Map<Character,Integer> x;
Map<Character,Integer> y;
for(int i=0;i<9;i++)
{
x=new HashMap<Character,Integer>();
y=new HashMap<Character,Integer>();
for(int j=0;j<9;j++)
{
if(board[i][j]>='0'&&board[i][j]<='9')
if(x.get(board[i][j])!=null)
return false;
else
x.put(board[i][j],1);
if(board[j][i]>='0'&&board[j][i]<='9')
if(y.get(board[j][i])!=null)
return false;
else
y.put(board[j][i],1);
}
x.clear();
y.clear();
}
if(isV(0,2,0,2,board)&&isV(3,5,0,2,board)&&isV(6,8,0,2,board)&&isV(0,2,3,5,board)&&isV(3,5,3,5,board)&&isV(6,8,3,5,board)&&isV(0,2,6,8,board)&&isV(3,5,6,8,board)&&isV(6,8,6,8,board))
return true;
else
return false;
}
}
