diff --git a/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md b/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md
index a4072e31..c19fb875 100644
--- a/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md
+++ b/website/content/ChapterFour/0400~0499/0417.Pacific-Atlantic-Water-Flow.md
@@ -95,5 +95,5 @@ func dfs(matrix [][]int, row, col int, visited *[][]bool, height int) {
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diff --git a/website/content/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md b/website/content/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md
new file mode 100644
index 00000000..505b3ab0
--- /dev/null
+++ b/website/content/ChapterFour/0400~0499/0419.Battleships-in-a-Board.md
@@ -0,0 +1,58 @@
+# [419. Battleships in a Board](https://leetcode.com/problems/battleships-in-a-board/)
+
+## 题目
+
+Given an `m x n` matrix `board` where each cell is a battleship `'X'` or empty `'.'`, return the number of the **battleships** on `board`.
+
+**Battleships** can only be placed horizontally or vertically on `board`. In other words, they can only be made of the shape `1 x k` (`1` row, `k` columns) or `k x 1` (`k` rows, `1` column), where `k` can be of any size. At least one horizontal or vertical cell separates between two battleships (i.e., there are no adjacent battleships).
+
+**Example 1:**
+
+
+
+```c
+Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
+Output: 2
+```
+
+**Example 2:**
+
+```c
+Input: board = [["."]]
+Output: 0
+```
+
+**Constraints:**
+
+- `m == board.length`
+- `n == board[i].length`
+- `1 <= m, n <= 200`
+- `board[i][j] is either '.' or 'X'`.
+
+**Follow up:** Could you do it in one-pass, using only `O(1)` extra memory and without modifying the values `board`?
+
+## 题目大意
+
+给定一个大小为`m × n`的矩阵 称之为甲板，矩阵单元格中的`'X'`表示战舰，`'.'`表示空位。
+
+战舰只能水平或竖直摆放在甲板上（换句话说，可以理解为联通的同一行`'X'`或同一列`'X'`只算作一个“战舰群”），任意俩个“战舰群”间都是不相邻的。返回甲板上“战舰群”的数量。
+
+## 解题思路
+
+题目进阶要求一次扫描算法，空间复杂度为`O(1)`，且不能修改矩阵中的值。
+
+因为题目中给定的两个“战舰群”间至少有一个水平或垂直的空位分隔，所以可以通过枚举每个战舰的左上顶点即可统计“战舰群”的个数。
+
+假设当前遍历到矩阵中`'X'`的位置为`(i, j)`，即 `board[i][j]='X'`。如果当前战舰属于一个新的“战舰群”，则需要满足以下条件：
+
+- 当前位置的上方位为空，即 `board[i-1][j]='.'`；
+- 当前位置的左方位为空，即 `board[i][j-1]='.'`；
+
+统计出所有左方位和上方位为空的战舰个数，即可得到“战舰群”的数量。
+
+
+----------------------------------------------
+
diff --git a/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md b/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md
index 05db370e..7abc8d85 100755
--- a/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md
+++ b/website/content/ChapterFour/0400~0499/0421.Maximum-XOR-of-Two-Numbers-in-an-Array.md
@@ -101,6 +101,6 @@ func findMaximumXOR1(nums []int) int {
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