问题：
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) orNx1(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:
...XXXXX...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?解决：
① 求战舰的个数，所谓的战舰就是只能是一行或者一列，不能有拐弯。本题限定了不会有相邻的两个战舰的存在，有了这一点限制，那么我们只需要遍历一次二维数组就行了，只要找出战舰的起始点。所谓的战舰起始点，就是指一条军舰上最左边的那个‘X’或者最上面的那个‘X’。
class Solution { //5ms
public int countBattleships(char[][] board) { if (board == null || board.length == 0 || board[0].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] == '.' || (i > 0 && board[i - 1][j] == 'X') || (j > 0 && board[i][j - 1] == 'X')){ continue; } count ++; } } return count; } }class Solution {//4ms
public int countBattleships(char[][] board) { if(board.length == 0) return 0; int count = 0; for(int i = 0; i < board.length; i ++) { for(int j = 0; j < board[0].length; j ++) { if(board[i][j] == 'X') { if(i > 0 && board[i - 1][j] == 'X') continue; if(j > 0 && board[i][j - 1] == 'X') continue; count ++; } } } return count; } }