419. Battleships in a Board
Medium
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:

Input: board = [["X",".",".","X"],[".",".",".","X"],[".",".",".","X"]]
Output:
2
Example 2:
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
?
func countBattleships(board [][]byte) int {
res := 0
for i:=0; i<len(board); i++ {
for j:=0; j<len(board[0]); j++ {
if board[i][j] == '.' { continue }
if i > 0 && board[i - 1][j] == 'X' { continue }
if j > 0 && board[i][j - 1] == 'X' { continue }
res++
}
}