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?
原来题目里已经说明不会包括 invalid 情况，想到判断的方法上去了。。。统计的话统计一个 battleship 左上的部分即可
public class Solution { public int countBattleships(char[][] board) { int m = board.length; if (m==0) return 0; int n = board[0].length; int count=0; for (int i=0; i<m; i++) { for (int j=0; j<n; j++) { if (board[i][j] == '.') continue; if (i > 0 && board[i-1][j] == 'X') continue; if (j > 0 && board[i][j-1] == 'X') continue; count++; } } return count; } }