419. Battleships in a Board

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:

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.

找到有多少条船，横向或纵向不相隔的X组成一条船，船之间至少有一个空点

解题思路：

遍历二维数组，每遍历到一个位置的时候，只需要考虑其左边和上面的位置是否为X即可

 1 int countBattleships(char** board, int boardRowSize, int boardColSize) {
 2     int i,j;
 3     int count=0;
 4     for(i=0;i<boardRowSize;i++)
 5         for(j=0;j<boardColSize;j++)
 6             if(board[i][j]==‘X‘)
 7             {
 8                 if(i==0)
 9                 {
10                     if(j==0||board[i][j-1]==‘.‘)
11                         count++;
12                 }
13                 else
14                 {
15                     if((j==0&&board[i-1][j]==‘.‘)||(board[i-1][j]==‘.‘&&board[i][j-1]==‘.‘))
16                     count++;
17                 }
18             }
19             return count;
20 }
21 //遍历到每个元素时候，只需要考虑其前面和上面是否为‘.‘即可

419. Battleships in a Board

原文：http://www.cnblogs.com/fcyworld/p/6146593.html