【LeetCode】36. Valid Sudoku

Difficulty: Medium

More:【目录】LeetCode Java实现

Description

https://leetcode.com/problems/valid-sudoku/

Determine if a 9x9 Sudoku board is valid. Only the filled cells need to be validated according to the following rules:

Each row must contain the digits 1-9 without repetition.
Each column must contain the digits 1-9 without repetition.
Each of the 9 3x3 sub-boxes of the grid must contain the digits 1-9 without repetition.

A partially filled sudoku which is valid.

The Sudoku board could be partially filled, where empty cells are filled with the character '.'.

Example 1:

Input:
[
  ["5","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
Output: true

Example 2:

Input:
[
  ["8","3",".",".","7",".",".",".","."],
  ["6",".",".","1","9","5",".",".","."],
  [".","9","8",".",".",".",".","6","."],
  ["8",".",".",".","6",".",".",".","3"],
  ["4",".",".","8",".","3",".",".","1"],
  ["7",".",".",".","2",".",".",".","6"],
  [".","6",".",".",".",".","2","8","."],
  [".",".",".","4","1","9",".",".","5"],
  [".",".",".",".","8",".",".","7","9"]
]
Output: false
Explanation: Same as Example 1, except with the 5 in the top left corner being 
    modified to 8. Since there are two 8's in the top left 3x3 sub-box, it is invalid.

Note:

A Sudoku board (partially filled) could be valid but is not necessarily solvable.
Only the filled cells need to be validated according to the mentioned rules.
The given board contain only digits 1-9 and the character '.'.
The given board size is always 9x9.

Intuition

Mehtod 1: Use three hashSet (rowSet, colSet, boxSet) to store digits in each row/ column/ sub-box.

Details:

　　i: 1 ->9 j: i ->9

　　rowSet: a[i][j]

　　colSet: a[j][i]

　　boxSet:

　　　　For a given sub-box, we can traverse a 3*3 sub-box using just j: a[ j/3 ] [ j%3 ]

　　　　We has 9 sub-boxes, so we can use i to switch in different sub-boxes: a[i/3*3+j/3][i%3*3+j%3]

Mehtod 2: Use only one hashSet to store a[i][j] in three ways.

　　Encode a[i][j] as a[i][j]+"r"+i to represent a[i][j] in row i;

　　Encode a[i][j] as a[i][j]+"c"+j to represent a[i][j] in colum j;

　　Encode a[i][j] as a[i][j]+"box"+i/3+j/3 to represent a[i][j] in box-i/3-j/3;

Solution

    //Method 1
    public boolean isValidSudoku(char[][] board) {
        for(int i=0; i<9; i++){
            HashSet<Character> rowSet = new HashSet<>();
            HashSet<Character> colSet = new HashSet<>();
            HashSet<Character> boxSet = new HashSet<>();
            for(int j=0; j<9; j++){
                if(board[i][j]!='.' && !rowSet.add(board[i][j]))
                    return false;
                if(board[j][i]!='.' && !colSet.add(board[j][i]))
                    return false;
                if(board[i/3*3+j/3][i%3*3+j%3]!='.' && !boxSet.add(board[i/3*3+j/3][i%3*3+j%3]))
                    return false;
            }
        }
        return true;
    }
    
    //Method 2: easier to understand
    public boolean isValidSudoku(char[][] board) {
        HashSet<String> set = new HashSet<>();
        for(int i=0; i<9; i++){
            for(int j=0; j<9; j++){
                if(board[i][j]!='.'){
                    String num = String.valueOf(board[i][j]);
                    if(!set.add(num+"r"+i) || !set.add(num+"c"+j) || !set.add(num+"box"+i/3+j/3))
                        return false;                    
                }
            }
        }
        return true;
    }

Complexity

Time complexity :

$Space complexity :$

What I've learned

1. Learn how to use i, j to represent elements in different rows and columns and sub-boxes.

2. For a given sub-box, we can traverse a 3*3 sub-box using just j: a[ j/3 ] [ j%3 ]

3. It's clever to store elements in a hashSet by convert elements to String.