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

 int countBattleships(char** board, int boardRowSize, int boardColSize) {
int i,j;
int count=;
for(i=;i<boardRowSize;i++)
for(j=;j<boardColSize;j++)
if(board[i][j]=='X')
{
if(i==)
{
if(j==||board[i][j-]=='.')
count++;
}
else
{
if((j==&&board[i-][j]=='.')||(board[i-][j]=='.'&&board[i][j-]=='.'))
count++;
}
}
return count;
}
//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