问题：
Given an 2D board, count how many different 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:
In the above board there are 2 battleships. Invalid Example:
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?
大意：
给出一个2D面板，计算其中有多少不同的战舰。战舰由‘X’表示，空地由‘.’表示。你可以假设满足下面的规则：
- 你会接受到一个有效的面板，只由战舰和空地组成。
- 战舰只能是水平的或者垂直的。也就是说，只能是 1xN（一行N列）或者 Nx1（N行一列）的形状，N可以是任何尺寸。
- 两个战舰之间至少有一个空地 - 没有相连的战舰。 例子：
上面的面板上有两艘战舰。 无效的例子：
这是一个无效的面板，你不会接受到 - 因为两艘战舰一定会有空白的点。 进阶： 你能不能使用O（1）的内存，并且不修改面板的值来完成？
思路：
这道题的情景很像我们玩的战船游戏，所以做起来也很有意思。
我们最好从左上角开始，遍历每个点，如果有X，就看它往右和往下有没有相邻的X，有的话也算做一艘战舰。已经检查过的点我们就不检查了，所以用一个二维数组来记录检查过的点。
代码（Java）：
代码语言：javascript复制public class Solution {
public int countBattleships(char[][] board) {
int result = 0;
// 用来记录有没有判断过
int[][] tempBoard = new int[board.length][board[0].length];
for (int i = 0; i < tempBoard.length; i ) {
for (int j = 0; j < tempBoard[0].length; j ) tempBoard[i][j] = 0;
}
for (int i = 0; i < board.length; i ) {
for (int j = 0; j < board[0].length; j ) {
if (tempBoard[i][j] == 0) {
if (board[i][j] - 'X' == 0) {
System.out.println(i " " j);
if (j 1 < board[0].length && board[i][j 1] - 'X' == 0) {// 竖向战舰
for (int k = j 1; k < board[0].length; k ) {
if (board[i][k] - 'X' == 0) tempBoard[i][k] = 1;
else break;
}
} else if (i 1 < board.length && board[i 1][j] - 'X' == 0) {// 横向战舰
for (int k = i 1; k < board.length; k ) {
if (board[k][j] - 'X' == 0) tempBoard[k][j] = 1;
else break;
}
}
tempBoard[i][j] = 1;
result ;
}
}
}
}
return result;
}
}他山之石：
代码语言：javascript复制 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;
}上面我的做法其实用了O（n）的内存，而且要进行多次循环，很耗时，这个就简单多了，每次遇到一个坐标，如果它既不是空地，他的上面和左边也没有X，那就说明这是一个新战舰，只记录这种新战舰的个数，很节省空间和时间。
合集：https://github.com/Cloudox/LeetCode-Record
