Solution 1:用set的简单循环
第一反应是set，set不能存重复的元素
Runtime: 21 ms, faster than 57.14% of Java online submissions for Valid Sudoku.
class Solution {
public boolean isValidSudoku(char[][] board) {
for(int i=0;i<9;i+=3){
for(int j=0;j<9;j+=3){
if(!CheckSquare(board,i,j)){
return false;
}
}
}
for(int i=0;i<9;i++){
if(!CheckRow(board,i)){
return false;
}
}
for(int j=0;j<9;j++){
if(!CheckColumn(board,j)){
return false;
}
}
return true;
}
private boolean CheckRow(char[][] board,int i){
Set<Character> set=new HashSet<>();
for(int j=0;j<9;j++){
if(board[i][j]!='.'){
if(!set.contains(board[i][j])){
set.add(board[i][j]);
}
else{
return false;
}
}
}
return true;
}
private boolean CheckColumn(char[][] board,int j){
Set<Character> set=new HashSet<>();
for(int i=0;i<9;i++){
if(board[i][j]!='.'){
if(!set.contains(board[i][j])){
set.add(board[i][j]);
}
else{
return false;
}
}
}
return true;
}
private boolean CheckSquare(char[][] board,int i,int j){
Set<Character> set=new HashSet<>();
int p=i,q=j;
for(p=i;p<i+3;p++){
for(q=j;q<j+3;q++){
if(board[p][q]!='.'){
if(!set.contains(board[p][q])){
set.add(board[p][q]);
}
else{
return false;
}
}
}
}
return true;
}
}
Solution 2:solution 1的再精简版本，把所有check函数整合成一个
public class Solution {
public boolean isValidSudoku(char[][] board) {
for (int i=0; i<9; i++) {
if (!isParticallyValid(board,i,0,i,8)) return false;
if (!isParticallyValid(board,0,i,8,i)) return false;
}
for (int i=0;i<3;i++){
for(int j=0;j<3;j++){
if (!isParticallyValid(board,i*3,j*3,i*3+2,j*3+2)) return false;
}
}
return true;
}
private boolean isParticallyValid(char[][] board, int x1, int y1,int x2,int y2){
Set singleSet = new HashSet();
for (int i= x1; i<=x2; i++){
for (int j=y1;j<=y2; j++){
if (board[i][j]!='.') if(!singleSet.add(board[i][j])) return false;
}
}
return true;
}
}
Solution 3: lc上genius!的办法，坚持了
public boolean isValidSudoku(char[][] board) {
Set seen = new HashSet();
for (int i=0; i<9; ++i) {
for (int j=0; j<9; ++j) {
char number = board[i][j];
if (number != '.')
if (!seen.add(number + " in row " + i) ||
!seen.add(number + " in column " + j) ||
!seen.add(number + " in block " + i/3 + "-" + j/3))
return false;
}
}
return true;
}