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  • Easy | LeetCode 36. 有效的数独

    36. 有效的数独

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    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。
    img

    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。

    解题思路

    方法一: 三次遍历

    像这样的题目, 数组本身元素数量有限, 即使直接暴力遍历, 也不会超时。

    方法二: 一次遍历

    使用 box_index = (row / 3) * 3 + columns / 3来描述某个位置(row, columns)的所在子数独的序号。

    image.png

    使用三个HashMap来跟踪当前所在行, 列, 方形内所有已经遇到的值。

    public boolean isValidSudoku(char[][] board) {
        // 初始化27个HaspMap, 分别是9行, 9列, 9个方格。
        HashMap<Integer, Integer>[] rows = new HashMap[9];
        HashMap<Integer, Integer> [] columns = new HashMap[9];
        HashMap<Integer, Integer> [] boxes = new HashMap[9];
        for (int i = 0; i < 9; i++) {
            rows[i] = new HashMap<Integer, Integer>();
            columns[i] = new HashMap<Integer, Integer>();
            boxes[i] = new HashMap<Integer, Integer>();
        }
    
        // validate a board
        for (int i = 0; i < 9; i++) {
            for (int j = 0; j < 9; j++) {
                char num = board[i][j];
                if (num != '.') {
                    int n = (int)num;
                    int box_index = (i / 3 ) * 3 + j / 3;
    				// 将当前元素加进HashMap当中
                    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);
    
                    // 发现添加元素进去过后, 个数大于1, 则直接返回false
                    if (rows[i].get(n) > 1 || columns[j].get(n) > 1 || boxes[box_index].get(n) > 1)
                        return false;
                }
            }
        }
    
        return true;
    }
    
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  • 原文地址:https://www.cnblogs.com/chenrj97/p/14456463.html
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