【算法】【线性表】【数组】有效的数独

1  题目

请你判断一个 9 x 9 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。（请参考示例图）

注意：

• 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。
• 空白格用 '.' 表示。

示例 1：
输入：board = 
[["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：board = 
[["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。

提示：

• board.length == 9
• board[i].length == 9
• board[i][j] 是一位数字（1-9）或者 '.'

2  解答

记住，别憨憨的去解题，比如空白的应该是多少，别去解，因为题目中说不一定有解的，只需要遍历围绕它的三个特征去遍历判断即可：

class Solution {
    public boolean isValidSudoku(char[][] board) {
        // 存放行的计数 表示第几行 0-8
        Map<Integer, Set<Integer>> rowMap = new HashMap();
        // 存放列的计数 key 表示第几列 0-8
        Map<Integer, Set<Integer>> columnMap = new HashMap();
        // 存放3*3的计数 
        Map<String, Set<Integer>> tableMap = new HashMap();

        for (int i=0; i<board.length; i++) {
            for (int j=0; j<board[i].length; j++) {
                char c = board[i][j];
                // 如果当前坐标下，数字为空略过
                if (c == '.') {
                    continue;
                }
                // 巧妙得到字符对应的数字
                int num = c - '0' + 1;
                // 放入列计数器
                Set<Integer> columnSet = columnMap.computeIfAbsent(j, (val) -> new HashSet<Integer>());
                // 当前列存在重复数字直接返回 false
                if (columnSet.contains(num)) {
                    return false;
                }
                columnSet.add(num);
                
                // 放入行计数器
                Set<Integer> rowSet = rowMap.computeIfAbsent(i, (val) -> new HashSet<Integer>());
                // 当前列存在重复数字直接返回 false
                if (rowSet.contains(num)) {
                    return false;
                }
                rowSet.add(num);
                
                // 放入九宫格计数器
                int tableI = i / 3;
                int tableJ = j / 3;
                Set<Integer> tableSet = tableMap.computeIfAbsent(String.format("%s-%s", tableI, tableJ), (val) -> new HashSet<Integer>());
                // 当前列存在重复数字直接返回 false
                if (tableSet.contains(num)) {
                    return false;
                }
                tableSet.add(num);
            }
        }
        // 遍历完都符合，返回 true
        return true;
    }
}

加油。