https://leetcode.com/problems/battleships-in-a-board/description/

Solution：

思路：
限定了不會(huì)有相鄰的兩個(gè)戰(zhàn)艦的存在，有了這一點(diǎn)限制，那么我們只需要遍歷一次二維數(shù)組就行了，只要找出戰(zhàn)艦的起始點(diǎn)。所謂的戰(zhàn)艦起始點(diǎn)，就是為X的點(diǎn)，而且該點(diǎn)的上方和左邊的點(diǎn)不能為X，所以我們只要找出所有滿足這個(gè)條件的點(diǎn)即可

Time Complexity: O(N) Space Complexity: O(N)

Solution Code：

class Solution {
    public int countBattleships(char[][] board) {
        int m = board.length;
        if (m==0) return 0;
        int n = board[0].length;
        
        int count=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;
                count++;
            }
        }
        
        return count;
    }
}