[leetcode-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?

 思路：

首先想到dfs。

void dfsbattle(vector<vector<char>>& board,vector<vector<bool>>& visited,int i,int j)
     {
         int m = board.size(), n = board[0].size();
         if (i < 0 || i >= m || j < 0 || j >= n || visited[i][j] || board[i][j] == '.')return;
         visited[i][j] = true;
         dfsbattle(board, visited, i + 1, j);
         dfsbattle(board, visited, i - 1, j);
         dfsbattle(board, visited, i , j+1);
         dfsbattle(board, visited, i , j-1);
     }
     int countBattleships(vector<vector<char>>& board)
     {
         if (board.empty())return 0;
         int m = board.size(), n = board[0].size();
         vector<vector<bool>>visited(m, vector<bool>(n, false));
         int ret = 0;
         for (int i = 0; i < m;i++)
         {
             for (int j = 0; j < n;j++)
             {
                 if (board[i][j]=='X'&& !visited[i][j])
                 {
                     dfsbattle(board, visited, i,j);
                     ret++;
                 }
             }
         }
         return ret;
     }

又题目提到不用额外空间，而且只遍历一次。。

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

参考：

https://discuss.leetcode.com/topic/64027/share-my-7-line-code-1-line-core-code-3ms-super-easy

 http://www.cnblogs.com/grandyang/p/5979207.html