LeetCode No.36 Valid Sudoku | #HashSet #Wrapper Classes

Q：

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 '.'.

Note:A valid Sudoku board (partially filled) is not necessarily solvable. Only the filled cells need to be validated.

A：

这个题目并不是要找数独unfilled区域的solution，而是要check每个filled cells是否valid。

public boolean isValidSudoku(char[][] board) {
    for(int i = 0; i<9; i++){
        HashSet rows = new HashSet();  //注释1
        HashSet columns = new HashSet();  //注释4
        HashSet cube = new HashSet();

        for (int j = 0; j < 9;j++){
            if(board[i][j]!='.' && !rows.add(board[i][j]))  //注释2
                return false;

            if(board[j][i]!='.' && !columns.add(board[j][i]))
                return false;

            int RowIndex = 3*(i/3);   //注释3
            int ColIndex = 3*(i%3);

            if(board[RowIndex + j/3][ColIndex + j%3]!='.' 
                   && !cube.add(board[RowIndex + j/3][ColIndex + j%3]))
                return false;
        }
    }
    return true;
}

注释1：

为什么类型里要用Character，不可以写成char？
正如Integer之于int，char是基本数据类型(primitive type)，Character是包装类型（Wrapper Classes，也就是类，实例化出来的叫对象），其中还涉及到自动封箱（Autoboxing），自动拆解（extracts）：

example 1:
int t = 10; Integer t1 = t; //自动封箱

Integer t = new Integer(10); int t1 = t ; //自动解封

example 2:
Interger = obj1; int num1 = 69; obj1 = num1; //自动封箱

Integer obj2 = new Integer (69); int num2; num2 = obj2; //自动解封

初次之外很重要的一点，Character作为char的包装类，作为一个类，它提供了很多方法。比如当我们使用HashSet.add()这个方法时，设计到object.equals()等方法，所以包装类的意义就体现出来了。

还有个情况，就是在数据类型不统一的时候，在操作数据的时候容易引起错误，报出“java.lang.ClassCastException”，比如Integer类型冲突了String，通过泛型，也可以进行一个限制，参考（http://www.cnblogs.com/lwbqqyumidi/p/3837629.html）。

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注释2：

先来看一下 java.util.HashSet.add() HashSet API方法：

public boolean add (E e)
Adds the specified element to this set if it is not already present. More formally, adds the specified element e to this set if this set contains no element e2 such that (e==null ? e2==null : e.equals(e2)). If this set already contains the element, the call leaves the set unchanged and returns false.
Specified by:
add in the interface Collection
add in interface Set
Overrides:
add in class AbstractCollection
Returns:
true if this set did not already contain the specified element

由此可知，!rows.add(board[i][j])表达的意思是当要加入新的数字时，如果不成功，说明matrix里已经存在了，那么这时出现了重复的值，要返回false，但！之后，返回true。如果那个cell的值也不是'.'，说明那个cell是有数字的。逻辑与（&&）操作之后，true && true，则判断条件成立，return false。

HashSet vs HashMap：
HashSet 实现了Set接口。不允许重复的值。 .add()
HashMap实现了Map接口。不允许重复的键。.put()

Map接口有两个基本的实现，HashMap和TreeMap。TreeMap保存了对象的排列次序，而HashMap则不能。HashMap允许键和值为null。HashMap是非synchronized的。

HashMap ，而我们这里参数进来其实不需要键值，使用Character封装类，匹配parameter char[][] board就好。

HashSet类维护了一个HashMap引用作为自己的成员变量：
public HashSet() { map = new HashMap(); }
HashSet源码分析：http://blog.chinaunix.net/uid-26864892-id-3167656.html

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注释3：

@jaqenhgar 解释了这里运算背后的逻辑：

0,0, 0,1, 0,2; < --- 3 Horizontal Steps followed by 1 Vertical step to next level.
1,0, 1,1, 1,2; < --- 3 Horizontal Steps followed by 1 Vertical step to next level.
2,0, 2,1, 2,2; < --- 3 Horizontal Steps.
And so on...But, the j iterates from 0 to 9.
But we need to stop after 3 horizontal steps, and go down 1 step vertical.

Use % for horizontal traversal. Because % increments by 1 for each j : 0%3 = 0 , 1%3 = 1, 2%3 = 2, and resets back. So this covers horizontal traversal for each block by 3 steps.
Use / for vertical traversal. Because / increments by 1 after every 3 j: 0/3 = 0; 1/3 = 0; 2/3 =0; 3/3 = 1.

So far, for a given block, you can traverse the whole block using just j.
But because j is just 0 to 9, it will stay only first block. But to increment block, use i. To move horizontally to next block, use % again : ColIndex = 3 * (i%3) (Multiply by 3 so that the next block is after 3 columns. Ie 0,0 is start of first block, second block is 0,3 (not0,1);
Similarly, to move to next block vertically, use / and multiply by 3 as explained above. Hope this helps.

逐个block traversal