## 419. Battleships in a Board

Given an 2D board, count how many battleships are in it. The battleships are represented with

`'X'`

s, empty slots are represented with `'.'`

s. You may assume the following rules:
- You receive a valid board, made of only battleships or empty slots.
- Battleships can only be placed horizontally or vertically. In other words, they can only be made of the shape
`1xN`

(1 row, N columns) or`Nx1`

(N rows, 1 column), where N can be of any size. - At least one horizontal or vertical cell separates between two battleships - there are no adjacent battleships.

**Example:**

X..X ...X ...XIn the above board there are 2 battleships.

**Invalid Example:**

...X XXXX ...XThis is an invalid board that you will not receive - as battleships will always have a cell separating between them.

**Follow up:**

Could you do it in **one-pass**, using only **O(1) extra memory** and **without modifying** the value of the board?

## Rust Solution

```
struct Solution;
impl Solution {
fn count_battleships(board: Vec<Vec<char>>) -> i32 {
let n = board.len();
let m = board[0].len();
let mut res = 0;
for i in 0..n {
for j in 0..m {
if Self::is_head(i, j, &board) {
res += 1;
}
}
}
res
}
fn is_head(i: usize, j: usize, board: &[Vec<char>]) -> bool {
if board[i][j] == '.' {
return false;
}
if i > 0 && board[i - 1][j] == 'X' {
return false;
}
if j > 0 && board[i][j - 1] == 'X' {
return false;
}
true
}
}
#[test]
fn test() {
let board = vec_vec_char![
['X', '.', '.', 'X'],
['.', '.', '.', 'X'],
['.', '.', '.', 'X']
];
let res = 2;
assert_eq!(Solution::count_battleships(board), res);
}
```

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