leetcode题解之36. 有效的数独

Posted 2020-12-12 刷题之路1

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。

思路

一个简单的解决方案是遍历该 9 x 9 数独 三 次，以确保：

• 行中没有重复的数字。
• 列中没有重复的数字。
• 3 x 3 子数独内没有重复的数字。

实际上，所有这一切都可以在一次迭代中完成。

---

方法：一次迭代

首先，让我们来讨论下面两个问题：

• 如何枚举子数独？

可以使用 box_index = (row / 3) * 3 + columns / 3，其中 / 是整数除法。

• 如何确保行 / 列 / 子数独中没有重复项？

可以利用 value -> count 哈希映射来跟踪所有已经遇到的值。

现在，我们完成了这个算法的所有准备工作：

• 遍历数独。
• 检查看到每个单元格值是否已经在当前的行 / 列 / 子数独中出现过：
  • 如果出现重复，返回 false。
  • 如果没有，则保留此值以进行进一步跟踪。
• 返回 true。

1 / 15

• Java
• Python

class Solution {
  public boolean isValidSudoku(char[][] board) {
    // init data
    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>();
    }
<span class="hljs-comment">// validate a board</span>
<span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> i = <span class="hljs-number">0</span>; i &lt; <span class="hljs-number">9</span>; i++) {
  <span class="hljs-keyword">for</span> (<span class="hljs-keyword">int</span> j = <span class="hljs-number">0</span>; j &lt; <span class="hljs-number">9</span>; j++) {
    <span class="hljs-keyword">char</span> num = board[i][j];
    <span class="hljs-keyword">if</span> (num != <span class="hljs-string">‘.‘</span>) {
      <span class="hljs-keyword">int</span> n = (<span class="hljs-keyword">int</span>)num;
      <span class="hljs-keyword">int</span> box_index = (i / <span class="hljs-number">3</span> ) * <span class="hljs-number">3</span> + j / <span class="hljs-number">3</span>;

      <span class="hljs-comment">// keep the current cell value</span>
      rows[i].put(n, rows[i].getOrDefault(n, <span class="hljs-number">0</span>) + <span class="hljs-number">1</span>);
      columns[j].put(n, columns[j].getOrDefault(n, <span class="hljs-number">0</span>) + <span class="hljs-number">1</span>);
      boxes[box_index].put(n, boxes[box_index].getOrDefault(n, <span class="hljs-number">0</span>) + <span class="hljs-number">1</span>);

      <span class="hljs-comment">// check if this value has been already seen before</span>
      <span class="hljs-keyword">if</span> (rows[i].get(n) &gt; <span class="hljs-number">1</span> || columns[j].get(n) &gt; <span class="hljs-number">1</span> || boxes[box_index].get(n) &gt; <span class="hljs-number">1</span>)
        <span class="hljs-keyword">return</span> <span class="hljs-keyword">false</span>;
    }
  }
}

<span class="hljs-keyword">return</span> <span class="hljs-keyword">true</span>;

}

}

class Solution:

def isValidSudoku(self, board):

"""

:type board: List[List[str]]

:rtype: bool

"""

# init data

rows = [{} for i in range(9)]

columns = [{} for i in range(9)]

boxes = [{} for i in range(9)]

    <span class="hljs-comment"># validate a board</span>
    <span class="hljs-keyword">for</span> i <span class="hljs-keyword">in</span> range(<span class="hljs-number">9</span>):
        <span class="hljs-keyword">for</span> j <span class="hljs-keyword">in</span> range(<span class="hljs-number">9</span>):
            num = board[i][j]
            <span class="hljs-keyword">if</span> num != <span class="hljs-string">‘.‘</span>:
                num = int(num)
                box_index = (i // <span class="hljs-number">3</span> ) * <span class="hljs-number">3</span> + j // <span class="hljs-number">3</span>
                
                <span class="hljs-comment"># keep the current cell value</span>
                rows[i][num] = rows[i].get(num, <span class="hljs-number">0</span>) + <span class="hljs-number">1</span>
                columns[j][num] = columns[j].get(num, <span class="hljs-number">0</span>) + <span class="hljs-number">1</span>
                boxes[box_index][num] = boxes[box_index].get(num, <span class="hljs-number">0</span>) + <span class="hljs-number">1</span>
                
                <span class="hljs-comment"># check if this value has been already seen before</span>
                <span class="hljs-keyword">if</span> rows[i][num] &gt; <span class="hljs-number">1</span> <span class="hljs-keyword">or</span> columns[j][num] &gt; <span class="hljs-number">1</span> <span class="hljs-keyword">or</span> boxes[box_index][num] &gt; <span class="hljs-number">1</span>:
                    <span class="hljs-keyword">return</span> <span class="hljs-literal">False</span>         
    <span class="hljs-keyword">return</span> <span class="hljs-literal">True</span>

复杂度分析

• 时间复杂度：$O(1)$，因为我们只对 81 个单元格进行了一次迭代。
• 空间复杂度：$O(1)$。

https://www.jianshu.com/p/ad0311d97760