求数独，只要求做出一个答案就可以。
刚开始对题意理解错误，以为答案是唯一的， 所以做了很久并没有做出来，发现答案不唯一之后，使用回溯。（还是借鉴了一下别人）
public class Solution {
public void solveSudoku(char[][] board) {
HashSet[] hashset = new HashSet[27];
for (int i = 0; i < 27; i++)
hashset[i] = new HashSet<Character>();
for (int i = 0; i < 9; i++) {
for (int j = 0; j < 9; j++) {
char Char = board[i][j];
if (Char != '.') {
hashset[i].add(Char);
hashset[9 + j].add(Char);
hashset[18 + (i / 3) * 3 + j / 3].add(Char);
}
}
}
int flag = 0;
char[][][] num = null ;
while ( flag == 0) {
flag = 1;
num = new char[9][9][9];
for (int i = 0; i < 9; i++) {// i代表第i个hashset
for (int j = 1; j < 10; j++) {// j代表1-9
char ch = (char) (j + '0');
int[] test = new int[2];
if (!hashset[i].contains(ch)) {
test[0] = 0;
for (int k = 0; k < 9; k++) {
char Ch = board[i][k];
if (Ch == '.') {
if (!hashset[9 + k].contains(ch) && !hashset[18 + (i / 3) * 3 + k / 3].contains(ch)) {
addNum(num, i, k, ch);
test[0]++;
test[1] = k;
}
}
}
}
if (test[0] == 1) {
board[i][test[1]] = ch;
hashset[i].add(ch);
flag = 0;
hashset[9 + test[1]].add(ch);
hashset[18 + (i / 3) * 3 + test[1] / 3].add(ch);
}
}
}
for (int qq = 0; qq < 9 && flag == 1; qq++) {
for (int j = 0; j < 9 && flag == 1; j++) {
if (getlen(num[qq][j]) == 1) {
char ch = num[qq][j][0];
board[qq][j] = ch;
flag = 0;
hashset[qq].add(ch);
hashset[9 + j].add(ch);
hashset[18 + (qq / 3) * 3 + j / 3].add(ch);
}
}
}
}
solve(board);
}
public boolean solve(char[][] board){
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == '.'){
for(char c = '1'; c <= '9'; c++){//trial. Try 1 through 9 for each cell
if(isValid(board, i, j, c)){
board[i][j] = c; //Put c for this cell
if(solve(board))
return true; //If it's the solution return true
else
board[i][j] = '.'; //Otherwise go back
}
}
return false;
}
}
}
return true;
}
public boolean isValid(char[][] board, int i, int j, char c){
for(int row = 0; row < 9; row++)
if(board[row][j] == c)
return false;
for(int col = 0; col < 9; col++)
if(board[i][col] == c)
return false;
for(int row = (i / 3) * 3; row < (i / 3) * 3 + 3; row++)
for(int col = (j / 3) * 3; col < (j / 3) * 3 + 3; col++)
if(board[row][col] == c)
return false;
return true;
}
public static int getlen(char[] num) {
int len = 0;
for (int i = 0; i < 9; i++) {
if (num[i] < '1' || num[i] > '9') {
return len;
} else
len++;
}
return len;
}
public static void addNum(char[][][] num, int num1, int num2, char ch) {
for (int i = 0; i < 9; i++) {
if (num[num1][num2][i] < '0' || num[num1][num2][i] > '9') {
num[num1][num2][i] = ch;
break;
}
}
}
}
回溯法还是比较简单的，就是在实现的时候，如果想要提高运行的速度和空间，那么需要费一些心思来考虑。
附上借鉴的代码
public class Solution {
public void solveSudoku(char[][] board) {
if(board == null || board.length == 0)
return;
solve(board);
}
public boolean solve(char[][] board){
for(int i = 0; i < board.length; i++){
for(int j = 0; j < board[0].length; j++){
if(board[i][j] == '.'){
for(char c = '1'; c <= '9'; c++){//trial. Try 1 through 9 for each cell
if(isValid(board, i, j, c)){
board[i][j] = c; //Put c for this cell
if(solve(board))
return true; //If it's the solution return true
else
board[i][j] = '.'; //Otherwise go back
}
}
return false;
}
}
}
return true;
}
public boolean isValid(char[][] board, int i, int j, char c){
//Check colum
for(int row = 0; row < 9; row++)
if(board[row][j] == c)
return false;
//Check row
for(int col = 0; col < 9; col++)
if(board[i][col] == c)
return false;
//Check 3 x 3 block
for(int row = (i / 3) * 3; row < (i / 3) * 3 + 3; row++)
for(int col = (j / 3) * 3; col < (j / 3) * 3 + 3; col++)
if(board[row][col] == c)
return false;
return true;
}
}