力扣 36. 有效的数独

题目说明

请你判断一个 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。（请参考示例图）
数独部分空格内已填入了数字，空白格用 ‘.’ 表示。

  • 提示：
    • 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。

示例

示例 1:

img

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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:

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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。

笔者理解

此题是一道数组算法问题，在力扣题库中被定义为中等题。

解法

当笔者阅读完此题后，直接按照横竖，九宫格来进行判断，让我们来看看具体如何实现的吧。

实现

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class Solution {
    /*
     * HashSet
     * 数组
     */
    char[][] board;
    public boolean isValidSudoku(char[][] board) {
        HashSet<Integer> set = new HashSet<>();
        this.board = board;

        // 横竖判断
        for (int i = 0; i < 9; i++) {
            boolean isConfirm = judge(i, 0, i + 1, 9);
            isConfirm &= judge(0, i, 9, i + 1);
            if (!isConfirm) {
                return false;
            }
        }

        // 九宫格判断
        for (int i = 0; i < 7; i+=3) {
            for (int j = 0; j < 7; j+=3) {
                if (!judge(i, j, i + 3, j + 3)) {
                    return false;
                }
            }
        }

        return true;
    }

    public boolean judge(int x, int y, int x0, int y0) {
        HashSet<Character> set = new HashSet<>();
        int tick = 0;
        int point = 0;

        // 除了 . 看有没有其他重复字符，有就说明数独无效
        for (int i = x; i < x0; i++) {
            for (int j = y; j < y0; j++) {
                char symbol = board[i][j];
                if (symbol == '.') {
                    point++;
                    continue;
                }
                if (!set.contains(symbol)) {
                    set.add(symbol);
                }
                else {
                    return false;
                }
                tick++;
            }
        }

        return true;
    }
}

时间和空间效率还行，可见此解法还比较适合此题。

image-20210904100811408

总结

本题是今天的一题，难度是为中等，感兴趣的朋友都可以去尝试一下，此题还有其他更多的解法，朋友们可以自己逐一尝试。