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.

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

Solution:

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        int ans = 0;
        
        int rows = board.size();
        if (rows == 0) {
            return 0;
        }
        int cols = board[0].size();
        
        for (int i = 0; i < rows; i++) {
            for (int j = 0; j < cols; j++) {
                if (board[i][j] == 'X') {
                    ans++;
                    if (check_right(board, i, j+1)) {
                        continue;
                    }
                    if (check_bottom(board, i+1, j)) {
                        continue;
                    }
                }
            }
        }
        return ans;
    }
    
    bool check_right(vector<vector<char>>& board, int i, int j) {
        if (j == board[0].size()) {
            return false;
        }
        bool flag = false;
        while (j < board[0].size() && board[i][j] == 'X') {
            board[i][j] = '.';
            j++;
            flag = true;
        }
        return flag;
    }
    bool check_bottom(vector<vector<char>>& board, int i, int j) {
        if (i == board.size()) {
            return false;
        }
        bool flag = false;
        while (i < board.size() && board[i][j] == 'X') {
            board[i][j] = '.';
            i++;
            flag = true;
        }
        return flag;
    }
};

======== One iteration and constant space and without modifying the board ====

int countBattleships(vector<vector<char>>& board) {
        if (board.empty() || board[0].empty()) return 0;
        int res = 0, m = board.size(), n = board[0].size();
        for (int i = 0; i < m; ++i) {
            for (int j = 0; j < n; ++j) {
                if (board[i][j] == '.' || (i > 0 && board[i - 1][j] == 'X') || (j > 0 && board[i][j - 1] == 'X')) continue;
                ++res;
            }
        }
        return res;
    }