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
?
public class Solution {
public int countBattleships(char[][] board) {
if (board == null || board.length == 0) {
return 0;
}
int count = 0;
int m = board.length;
int n = board[0].length;
for (int i = 0; i < m; i++) {
for (int j = 0; j < n; j++) {
if (board[i][j] != '.'
&& (j == 0 || board[i][j - 1] != 'X')
&& (i <= 0 || board[i - 1][j] != 'X')) {
count++;
}
}
}
return count;
}
}