Description#
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:
![](https://spcdn.pages.dev/leetcode/problems/0419.Battleships%20in%20a%20Board/images/battelship-grid.jpg)
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
?
Solutions#
Solution 1#
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| class Solution:
def countBattleships(self, board: List[List[str]]) -> int:
m, n = len(board), len(board[0])
ans = 0
for i in range(m):
for j in range(n):
if board[i][j] == '.':
continue
if i > 0 and board[i - 1][j] == 'X':
continue
if j > 0 and board[i][j - 1] == 'X':
continue
ans += 1
return ans
|
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| class Solution {
public int countBattleships(char[][] board) {
int m = board.length, n = board[0].length;
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == '.') {
continue;
}
if (i > 0 && board[i - 1][j] == 'X') {
continue;
}
if (j > 0 && board[i][j - 1] == 'X') {
continue;
}
++ans;
}
}
return ans;
}
}
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| class Solution {
public:
int countBattleships(vector<vector<char>>& board) {
int m = board.size(), n = board[0].size();
int ans = 0;
for (int i = 0; i < m; ++i) {
for (int j = 0; j < n; ++j) {
if (board[i][j] == '.') continue;
if (i > 0 && board[i - 1][j] == 'X') continue;
if (j > 0 && board[i][j - 1] == 'X') continue;
++ans;
}
}
return ans;
}
};
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| func countBattleships(board [][]byte) int {
m, n := len(board), len(board[0])
ans := 0
for i := 0; i < m; i++ {
for j := 0; j < n; j++ {
if board[i][j] == '.' {
continue
}
if i > 0 && board[i-1][j] == 'X' {
continue
}
if j > 0 && board[i][j-1] == 'X' {
continue
}
ans++
}
}
return ans
}
|