Merge pull request #423 from LetMeFly666/419

添加问题“419.甲板上的战舰”的代码和题解

Author: LetMeFly666

Codes/0419-battleships-in-a-board_20240611.cpp

@@ -0,0 +1,24 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-06-11 10:15:35
+ * @LastEditors: LetMeFly
+ * @LastEditTime: 2024-06-11 12:19:59
+ */
+#ifdef _WIN32
+#include "_[1,2]toVector.h"
+#endif
+
+class Solution {
+public:
+    int countBattleships(vector<vector<char>>& board) {
+        int ans = 0;
+        for (int i = 0; i < board.size(); i++) {
+            for (int j = 0; j < board[0].size(); j++) {
+                if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.')) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+};

Codes/0419-battleships-in-a-board_20240611.go

@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-06-11 12:24:27
+ * @LastEditors: LetMeFly
+ * @LastEditTime: 2024-06-11 12:26:00
+ */
+package main
+
+func countBattleships(board [][]byte) int {
+    ans := 0
+    for i := 0; i < len(board); i++ {
+        for j := 0; j < len(board[0]); j++ {
+            if board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.') {
+                ans++
+            }
+        }
+    }
+    return ans
+}

Codes/0419-battleships-in-a-board_20240611.java

@@ -0,0 +1,19 @@
+/*
+ * @Author: LetMeFly
+ * @Date: 2024-06-11 12:22:38
+ * @LastEditors: LetMeFly
+ * @LastEditTime: 2024-06-11 12:24:00
+ */
+class Solution {
+    public int countBattleships(char[][] board) {
+        int ans = 0;
+        for (int i = 0; i < board.length; i++) {
+            for (int j = 0; j < board[0].length; j++) {
+                if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.')) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+}

Codes/0419-battleships-in-a-board_20240611.py

@@ -0,0 +1,11 @@
+'''
+Author: LetMeFly
+Date: 2024-06-11 12:20:29
+LastEditors: LetMeFly
+LastEditTime: 2024-06-11 12:21:50
+'''
+from typing import List
+
+class Solution:
+    def countBattleships(self, board: List[List[str]]) -> int:
+        return sum(board[i][j] == 'X' and (i == 0 or board[i - 1][j] == '.') and (j == 0 or board[i][j - 1] == '.') for j in range(len(board[0])) for i in range(len(board)))

README.md

@@ -174,6 +174,7 @@
 |0409.最长回文串|简单|<a href="https://leetcode.cn/problems/longest-palindrome/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/17/LeetCode%200409.%E6%9C%80%E9%95%BF%E5%9B%9E%E6%96%87%E4%B8%B2/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127359550" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/longest-palindrome/solution/letmefly-409zui-chang-hui-wen-chuan-by-t-xsid/" target="_blank">LeetCode题解</a>|
 |0410.分割数组的最大值|困难|<a href="https://leetcode.cn/problems/split-array-largest-sum/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/01/21/LeetCode%200410.%E5%88%86%E5%89%B2%E6%95%B0%E7%BB%84%E7%9A%84%E6%9C%80%E5%A4%A7%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/135728821" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/split-array-largest-sum/solutions/2613456/letmefly-410fen-ge-shu-zu-de-zui-da-zhi-agh2v/" target="_blank">LeetCode题解</a>|
 |0415.字符串相加|简单|<a href="https://leetcode.cn/problems/add-strings/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2023/07/17/LeetCode%200415.%E5%AD%97%E7%AC%A6%E4%B8%B2%E7%9B%B8%E5%8A%A0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/131758488" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/add-strings/solutions/2346886/letmefly-415zi-fu-chuan-xiang-jia-mo-ni-r1ph3/" target="_blank">LeetCode题解</a>|
+|0419.甲板上的战舰|中等|<a href="https://leetcode.cn/problems/battleships-in-a-board/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/06/11/LeetCode%200419.%E7%94%B2%E6%9D%BF%E4%B8%8A%E7%9A%84%E6%88%98%E8%88%B0/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/139595909" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/battleships-in-a-board/solutions/2807404/letmefly-419jia-ban-shang-de-zhan-jian-t-2bhw/" target="_blank">LeetCode题解</a>|
 |0421.数组中两个数的最大异或值|中等|<a href="https://leetcode.cn/problems/maximum-xor-of-two-numbers-in-an-array/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/19/LeetCode%200421.%E6%95%B0%E7%BB%84%E4%B8%AD%E4%B8%A4%E4%B8%AA%E6%95%B0%E7%9A%84%E6%9C%80%E5%A4%A7%E5%BC%82%E6%88%96%E5%80%BC/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127413219" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/maximum-xor-of-two-numbers-in-an-array/solution/letmefly-421shu-zu-zhong-liang-ge-shu-de-zaxe/" target="_blank">LeetCode题解</a>|
 |0429.N叉树的层序遍历|中等|<a href="https://leetcode.cn/problems/n-ary-tree-level-order-traversal/solutions/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2024/02/17/LeetCode%200429.N%E5%8F%89%E6%A0%91%E7%9A%84%E5%B1%82%E5%BA%8F%E9%81%8D%E5%8E%86/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/136136336" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/n-ary-tree-level-order-traversal/solutions/2642557/letmefly-429n-cha-shu-de-ceng-xu-bian-li-k66j/" target="_blank">LeetCode题解</a>|
 |0445.两数相加II|中等|<a href="https://leetcode.cn/problems/add-two-numbers-ii/" target="_blank">题目地址</a>|<a href="https://blog.letmefly.xyz/2022/10/14/LeetCode%200445.%E4%B8%A4%E6%95%B0%E7%9B%B8%E5%8A%A0II/" target="_blank">题解地址</a>|<a href="https://letmefly.blog.csdn.net/article/details/127316269" target="_blank">CSDN题解</a>|<a href="https://leetcode.cn/problems/add-two-numbers-ii/solution/letmefly-445liang-shu-xiang-jia-ii-by-ti-vbh5/" target="_blank">LeetCode题解</a>|

Solutions/LeetCode 0419.甲板上的战舰.md

@@ -0,0 +1,131 @@
+---
+title: 419.甲板上的战舰
+date: 2024-06-11 12:27:23
+tags: [题解, LeetCode, 中等, 深度优先搜索, 数组, 矩阵]
+---
+
+# 【LetMeFly】419.甲板上的战舰：统计战舰头（左上角）——附Py一行版
+
+力扣题目链接：[https://leetcode.cn/problems/battleships-in-a-board/](https://leetcode.cn/problems/battleships-in-a-board/)
+
+<p>给你一个大小为 <code>m x n</code> 的矩阵 <code>board</code> 表示甲板，其中，每个单元格可以是一艘战舰 <code>'X'</code> 或者是一个空位 <code>'.'</code> ，返回在甲板 <code>board</code> 上放置的 <strong>战舰</strong> 的数量。</p>
+
+<p><strong>战舰</strong> 只能水平或者垂直放置在 <code>board</code> 上。换句话说，战舰只能按 <code>1 x k</code>（<code>1</code> 行，<code>k</code> 列）或 <code>k x 1</code>（<code>k</code> 行，<code>1</code> 列）的形状建造，其中 <code>k</code> 可以是任意大小。两艘战舰之间至少有一个水平或垂直的空位分隔 （即没有相邻的战舰）。</p>
+
+<p>&nbsp;</p>
+
+<p><strong>示例 1：</strong></p>
+<img alt="" src="https://assets.leetcode.com/uploads/2021/04/10/battelship-grid.jpg" style="width: 333px; height: 333px;" />
+<pre>
+<strong>输入：</strong>board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
+<strong>输出：</strong>2
+</pre>
+
+<p><strong>示例 2：</strong></p>
+
+<pre>
+<strong>输入：</strong>board = [["."]]
+<strong>输出：</strong>0
+</pre>
+
+<p>&nbsp;</p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>m == board.length</code></li>
+	<li><code>n == board[i].length</code></li>
+	<li><code>1 &lt;= m, n &lt;= 200</code></li>
+	<li><code>board[i][j]</code> 是 <code>'.'</code> 或 <code>'X'</code></li>
+</ul>
+
+<p>&nbsp;</p>
+
+<p><strong>进阶：</strong>你可以实现一次扫描算法，并只使用<strong> </strong><code>O(1)</code><strong> </strong>额外空间，并且不修改 <code>board</code> 的值来解决这个问题吗？</p>
+
+
+    
+## 解题方法：统计战舰左上角
+
+这题的意思是，若出现战舰```X```，则战舰```X```的形状必定是“一行或一列连着几个”。
+
+因此我们直接统计有多少个“战舰头”不就可以了吗。
+
+遍历每个方格，若当前方格为战舰```X```，则满足以下条件时该方格为“战舰头”：
+
+> + 该方格左边是边界或空地
+> + 该方格上边是边界或空地
+
++ 时间复杂度$O(size(board))$
++ 空间复杂度$O(1)$
+
+### AC代码
+
+#### C++
+
+```cpp
+class Solution {
+public:
+    int countBattleships(vector<vector<char>>& board) {
+        int ans = 0;
+        for (int i = 0; i < board.size(); i++) {
+            for (int j = 0; j < board[0].size(); j++) {
+                if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.')) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+};
+```
+
+#### Go
+
+```go
+// package main
+
+func countBattleships(board [][]byte) int {
+    ans := 0
+    for i := 0; i < len(board); i++ {
+        for j := 0; j < len(board[0]); j++ {
+            if board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.') {
+                ans++
+            }
+        }
+    }
+    return ans
+}
+```
+
+#### Java
+
+```java
+class Solution {
+    public int countBattleships(char[][] board) {
+        int ans = 0;
+        for (int i = 0; i < board.length; i++) {
+            for (int j = 0; j < board[0].length; j++) {
+                if (board[i][j] == 'X' && (i == 0 || board[i - 1][j] == '.') && (j == 0 || board[i][j - 1] == '.')) {
+                    ans++;
+                }
+            }
+        }
+        return ans;
+    }
+}
+```
+
+#### Python
+
+```python
+# from typing import List
+
+class Solution:
+    def countBattleships(self, board: List[List[str]]) -> int:
+        return sum(board[i][j] == 'X' and (i == 0 or board[i - 1][j] == '.') and (j == 0 or board[i][j - 1] == '.') for j in range(len(board[0])) for i in range(len(board)))
+```
+
+> 同步发文于CSDN和我的[个人博客](https://blog.letmefly.xyz/)，原创不易，转载经作者同意后请附上[原文链接](https://blog.letmefly.xyz/2024/06/11/LeetCode%200419.%E7%94%B2%E6%9D%BF%E4%B8%8A%E7%9A%84%E6%88%98%E8%88%B0/)哦~
+>
+> Tisfy：[https://letmefly.blog.csdn.net/article/details/139595909](https://letmefly.blog.csdn.net/article/details/139595909)