[leetcode]Valid Sudoku
来源：程序员人生 发布时间：2014-11-11 08:19:19 阅读次数：2197次
问题描写：
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.
Note:
A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.
代码：
public class Valid_Sudoku { //java
public boolean isValidSudoku(char[][] board) {
int size = 9;
int [] member = new int[size]; //record if is occur;
//valid row
for(int i = 0; i < 9; i++){
for(int k = 0; k < 9; k++)
member[k] = 0;
for(int j = 0; j < 9; j++){
if(board[i][j] == '.')
continue;
int pos = board[i][j]-'0';
if(member[pos⑴] == 1)
return false;
else member[pos⑴] = 1;
}
}
//valid col
for(int i = 0; i < 9; i++){
for(int k = 0; k < 9; k++)
member[k] = 0;
for(int j = 0; j < 9; j++){
if(board[j][i] == '.')
continue;
int pos = board[j][i]-'0';
if(member[pos⑴] == 1)
return false;
else member[pos⑴] = 1;
}
}
//valid cube
for(int ibegin = 0; ibegin < 9; ibegin = ibegin+3){
for(int jbegin = 0; jbegin < 9; jbegin = jbegin+3){
for(int k = 0; k < 9; k++)
member[k] = 0;
for(int i = ibegin; i < ibegin+3; i++){
for(int j = jbegin; j < jbegin+3; j++){
if(board[i][j] == '.')
continue;
int pos = board[i][j]-'0';
if(member[pos⑴] == 1)
return false;
else member[pos⑴] = 1;
}
}
}
}
return true;
}
public static void main(String [] args){
String[] boardStr = {"......5..",
".........",
".........",
"93..2.4..",
"..7...3..",
".........",
"...34....",
".....3...",
".....52.."};
char [][] board = new char [9][9];
for(int i =0; i< boardStr.length; i++){
for(int j = 0; j<boardStr[i].length(); j++){
board[i][j] = boardStr[i].charAt(j);
}
}
Valid_Sudoku vs = new Valid_Sudoku();
System.out.println(vs.isValidSudoku(board));
}
}
生活不易，码农辛苦
如果您觉得本网站对您的学习有所帮助,可以手机扫描二维码进行捐赠
