JAVA程序设计：有效的数独（LeetCode：36）

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。

数字 1-9 在每一行只能出现一次。
数字 1-9 在每一列只能出现一次。
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

思路：一个简单的解决方案是遍历该 9 x 9 三次。我们可以采用hashmap将遍历缩减到一次。

class Solution {
    public boolean isValidSudoku(char[][] board) {
        HashMap[] rows=new HashMap[9];
        HashMap[] columns=new HashMap[9];
        HashMap[] boxes=new HashMap[9];
        for(int i=0;i<9;i++)
        {
        	rows[i]=new HashMap();
        	columns[i]=new HashMap();
        	boxes[i]=new HashMap();
        }
        for(int i=0;i<9;i++)
        	for(int j=0;j<9;j++)
        	{
        		char num=board[i][j];
        		if(num=='.')
        			continue;
        		int n=(int)num;
        		int box_index=(i/3)*3+j/3;
        		rows[i].put(n, rows[i].getOrDefault(n, 0)+1);
        		columns[j].put(n, columns[j].getOrDefault(n, 0)+1);
        		boxes[box_index].put(n, boxes[box_index].getOrDefault(n, 0)+1);
        		if(rows[i].get(n)>1 || columns[j].get(n)>1 || boxes[box_index].get(n)>1)
        			return false;
        	}
        return true;
    }
}