题解——Leetcode 419. Battleships in a Board 难度：Medium

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 shape1xN (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.

Could you do it in one-pass, using onlyO(1) extra memory andwithout modifying the value of the board?

题目如上，简化后的题意如下：

给你一个方形的板，上面分布着战舰和狭槽，其中战舰用‘X’表示，狭槽用‘.’表示，但战舰的位置有一定限制。战舰可以是一行也可以是一列，而且战舰之间必须隔着狭槽。通过编程算出板上战舰的数量。

除此之外，题目还要求程序只能对板进行一次遍历，使用的额外内存的空间复杂度为O（1），并且不能修改板上战舰和狭槽的分布。

C++程序如下：

    class Solution {      public:          int countBattleships(vector<vector<char>>& board) {              int count=0;              for(int i = 0; i < board.size(); i++){                  for(int j = 0; j < board[0].size(); j++)                      if(board[i][j]=='X'&&(i==0 || board[i-1][j]!='X')&&(j==0 || board[i][j-1] != 'X'))                          count++;              }              return count;          }      };  

战舰的数量等于战舰头的数量，所以当遍历board找到一个'X'时只需要判断它是不是战舰头即可。相对于战舰的非头位置，战舰横向排列时战舰头的左边必定为'.'，纵向排列时上边必定为'.'。

函数countBattleships()的返回值为战舰的数量，函数参数为数据类型为二维容器的board，由于需要遍历board,所以设置两层嵌套的循环，外层循环次数为board.size()即板的行数，内层循环次数为board[0].size()即板的列数。在循环内判断遍历到的元素是否为战舰头，如果是就将战舰数量加1。