Class Solution
java.lang.Object
g0401_0500.s0419_battleships_in_a_board.Solution
419 - Battleships in a Board.<p>Medium</p>
<p>Given an <code>m x n</code> matrix <code>board</code> where each cell is a battleship <code>'X'</code> or empty <code>'.'</code>, return <em>the number of the <strong>battleships</strong> on</em> <code>board</code>.</p>
<p><strong>Battleships</strong> can only be placed horizontally or vertically on <code>board</code>. In other words, they can only be made of the shape <code>1 x k</code> (<code>1</code> row, <code>k</code> columns) or <code>k x 1</code> (<code>k</code> rows, <code>1</code> column), where <code>k</code> can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).</p>
<p><strong>Example 1:</strong></p>
<p><img src="https://assets.leetcode.com/uploads/2021/04/10/battelship-grid.jpg" alt="" /></p>
<p><strong>Input:</strong> board = [[“X”,“.”,“.”,“X”],[“.”,“.”,“.”,“X”],[“.”,“.”,“.”,“X”]]</p>
<p><strong>Output:</strong> 2</p>
<p><strong>Example 2:</strong></p>
<p><strong>Input:</strong> board = [[“.”]]</p>
<p><strong>Output:</strong> 0</p>
<p><strong>Constraints:</strong></p>
<ul>
<li><code>m == board.length</code></li>
<li><code>n == board[i].length</code></li>
<li><code>1 <= m, n <= 200</code></li>
<li><code>board[i][j]</code> is either <code>'.'</code> or <code>'X'</code>.</li>
</ul>
<p><strong>Follow up:</strong> Could you do it in one-pass, using only <code>O(1)</code> extra memory and without modifying the values <code>board</code>?</p>
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Solution
public Solution()
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Method Details
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countBattleships
public int countBattleships(char[][] board)
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