## 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
...X
```
In the above board there are 2 battleships.

Invalid Example:

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

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 {
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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