LeetCode

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