419. Battleships in a Board

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) 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.

Follow up:
Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

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;
    }
};

419. Battleships in a Board

原文：http://www.cnblogs.com/wujufengyun/p/7462424.html