LeetCode | Valid Sudoku

题目

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.

分析

根据数独规则，判断行、列、子矩阵

代码

import java.util.HashSet;

public class ValidSudoku {
	public boolean isValidSudoku(char[][] board) {
		int N = board.length;
		int subSize = (int) Math.sqrt(N);
		// row
		for (int i = 0; i < N; ++i) {
			HashSet<Character> set = new HashSet<Character>();
			for (int j = 0; j < N; ++j) {
				char c = board[i][j];
				if (c != '.') {
					if (set.contains(c)) {
						return false;
					}
					set.add(c);
				}
			}
		}

		// column
		for (int j = 0; j < N; ++j) {
			HashSet<Character> set = new HashSet<Character>();
			for (int i = 0; i < N; ++i) {
				char c = board[i][j];
				if (c != '.') {
					if (set.contains(c)) {
						return false;
					}
					set.add(c);
				}
			}
		}

		// subboard
		for (int i = 0; i < N; ++i) {
			int originRow = (i / subSize) * subSize;
			int originColumn = (i % subSize) * subSize;
			HashSet<Character> set = new HashSet<Character>();
			for (int j = 0; j < N; ++j) {
				int row = originRow + j / subSize;
				int column = originColumn + j % subSize;
				char c = board[row][column];
				if (c != '.') {
					if (set.contains(c)) {
						return false;
					}
					set.add(c);
				}
			}
		}
		return true;
	}
}