LeetCode Valid Sudoku

原题链接在这里：https://leetcode.com/problems/valid-sudoku/

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.

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(H) Sudoku Solver

这道题基本没有什么快速方法，唯一就是brute force.

先一行一行检查

然后一列一列检查，这里注意外层for loop是j

最后一个一个小方块来检查，关键是如何检查sub box. 这里设值index 0-8 box, loop row with box/3*3 - box/3*3+3, loop column with box%3*3 - box%3*3+3.

Time Complexity: O(n^2). Space: O(1).

AC Java:

public class Solution {
public boolean isValidSudoku(char[][] board) {
if(board == null || board.length != 9 || board[0].length != 9){
return false;
}

HashSet<Character> hs = new HashSet<Character>();
//check each row
for(int i = 0; i<board.length; i++){
hs.clear();
for(int j = 0; j<board[0].length; j++){
if(board[i][j] != '.'){
if(!hs.contains(board[i][j])){
hs.add(board[i][j]);
}else{
return false;
}
}
}
}

//check each column
for(int j = 0; j<board[0].length; j++){
hs.clear();
for(int i = 0; i<board.length; i++){
if(board[i][j] != '.'){
if(!hs.contains(board[i][j])){
hs.add(board[i][j]);
}else{
return false;
}
}
}
}

//check each subbox
for(int box = 0; box<9; box++){
hs.clear();
for(int i = box/3*3; i<box/3*3+3; i++){
for(int j = box%3*3; j<box%3*3+3; j++){
if(board[i][j] != '.'){
if(!hs.contains(board[i][j])){
hs.add(board[i][j]);
}else{
return false;
}
}
}
}
}
return true;
}
}

后面跟上Sudoku Solver.