leetcode36-Valid Sudoku有效数独

判断一个 9x9 的数独是否有效。只需要根据以下规则，验证已经填入的数字是否有效即可。

1. 数字 1-9 在每一行只能出现一次。
2. 数字 1-9 在每一列只能出现一次。
3. 数字 1-9 在每一个以粗实线分隔的 3x3 宫内只能出现一次。

上图是一个部分填充的有效的数独。

数独部分空格内已填入了数字，空白格用 '.' 表示。

示例 1:

输入:
[
  ["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"]
]
输出: true

示例 2:

输入:
[
  ["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"]
]
输出: false
解释: 除了第一行的第一个数字从 5 改为 8 以外，空格内其他数字均与 示例1 相同。
     但由于位于左上角的 3x3 宫内有两个 8 存在, 因此这个数独是无效的。

说明:

• 一个有效的数独（部分已被填充）不一定是可解的。
• 只需要根据以上规则，验证已经填入的数字是否有效即可。
• 给定数独序列只包含数字 1-9 和字符 '.' 。

• 给定数独永远是 9x9 形式的。

看了很多网上的解答没发现有关于位运算的解法详细解释，只有赤果果的答案摆在那，这篇文章主要详细解释位运算的过程。在位运算之前我还发现一个很好理解的解题方案直接使用Set来存储每一行每一列每一区块数字的位置就是下边这个啦。

public boolean isValidSudoku(char[][] board) {
	    Set seen = new HashSet();
	    for (int i=0; i<9; ++i) {
	        for (int j=0; j<9; ++j) {
	            char number = board[i][j];
	            if (number != '.')
	                if (!seen.add(number + " in row " + i) ||
	                    !seen.add(number + " in column " + j) ||
	                    !seen.add(number + " in block " + i/3 + "-" + j/3))
	                    return false;
	        }
	    }
	    return true;
	}

这个方案还是比较好理解的，给每个元素都固定位置，一旦产生重复则false。使用这个方案提交leetcode的时候发现用时很长，大约打败22%的人。本着对算法运行速度的极致追求，然后 额嗯，。就点开了运行最快的解决方案的答案，发现不大好理解，随即去百度等各大搜索网站去搜寻。最终没有任何发现。

下边就详细介绍下这个方案的每一个细节

首先也是循环每个元素节点，然后通过

int a = board[i][j] -'0';

转换成int类型，答案中写的是减1。始终不明白，后来在去掉不减的时候debug发现得到的a并不是原数，而是原数的Ascii码。岁而明白这是char转int的方法。

int n = 1 << a;  

这个就是正常的按位左移运算了，主要用来判断是否有相同的元素存在。

if ((signs[0][i] & n) != 0 
                		|| (signs[1][j] & n) != 0 
                		|| (signs[2][cubeIndex] & n) != 0){
                    return false;
                }

这段我认为是最难理解的地方，我们可以把它拆分来看，先看每行，其他先不关注。代码中使用了&，并且上边是按元素左位移的，所以只有当元素相等的情况下值才不回为0，&为相对应位都是1，则结果为1，否则为0。每一列、每一区块的判断逻辑都是如此。

signs[0][i] |= n;
                signs[1][j] |= n;
                signs[2][cubeIndex] |= n;

这段就是把循环的元素都合并到一起，这样才能保证在上边进行重复判断的时候所有的元素都会判断上。下面附上这种方案的完整解决方案：

public boolean isValidSudoku1(char[][] board) {
        int[][] signs = new int[3][9];
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
            	
                if (board[i][j] == '.'){
                    continue;
                }
                int a = board[i][j] -'0';   //char 转  int方法
                int n = 1 << a;             
                int cubeIndex = i / 3 * 3 + j / 3;
                if ((signs[0][i] & n) != 0 
                		|| (signs[1][j] & n) != 0 
                		|| (signs[2][cubeIndex] & n) != 0){
                    return false;
                }
                
                signs[0][i] |= n;
                signs[1][j] |= n;
                signs[2][cubeIndex] |= n;
            }
        }
        return true;
    }