Valid Sudoku @LeetCode

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.

	package leetcode;
	import java.util.ArrayList;
	/**
	* Solution: Don't need to care anything but that three Sodoku rules.
	* 1. Each row must have the number 1-9 occuring just once.
	* 2. Each column .....
	* 3. The number 1-9 must occur just once in each of sub-boxes of the grid.
	* Althought I think there's something I need to care, something I may miss. But that's the result.
	* I refactor sub-box check to avoid another nest-loop.
	*
	* Reference: http://www.cnblogs.com/zhaolizhen/p/Sudoku.html
	* @author jeffwan
	* @date Apr 3, 2014
	*/
	public class IsValidSudoku {
	public boolean isValidSudoku(char[][] board) {
	if (board == null || board.length == 0 || board[0].length == 0) {
	return false;
	}
	int length = board.length;
	ArrayList<Integer> visited = new ArrayList<Integer>();
	// Check every row
	for (int i = 0; i < length; i++) {
	visited.clear();
	for (int j = 0; j < length; j++) {
	if (!process(visited, board[i][j])) {
	return false;
	}
	}
	}
	// Check every colum
	for (int i = 0; i < length; i++) {
	visited.clear();
	for (int j = 0; j < length; j++) {
	if (!process(visited, board[j][i])) {
	return false;
	}
	}
	}
	// Check every subx-box
	for (int i = 0; i < length; i += 3) {
	for (int j = 0; j < length; j += 3) {
	visited.clear();
	for (int k = 0; k < length; k++) {
	if (!process(visited, board[i + k / 3][j + k % 3])) {
	return false;
	}
	}
	}
	}
	return true;
	}
	private boolean process(ArrayList<Integer> visited, char digit) {
	if (digit == '.') {
	return true;
	}
	int num = Character.getNumericValue(digit);
	if (num < 1 || num > 9 || visited.contains(num)) {
	return false;
	}
	visited.add(num);
	return true;
	}
	}

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