leetcode 419 Battleships in a Board
z
 2019-04-18 19:03:52  

Given an 2D board, count how many battleships are in it. The battleships are represented with

1
'X'

s, empty slots are represented with

1
'.'

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:

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2
3
X..X
...X
...X

In the above board there are 2 battleships.

Invalid Example:

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2
3
...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?

  • 每当遇到’x’的时候，判断它是不是它所在的battership的最左上角的那个，如果是的话，count ++
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class Solution {
    public int countBattleships(char[][] board) {
        int count = 0;
        if (board == null || board.length == 0 || board[0].length == 0)
            return count;
        for (int i=0; i<board.length; i++) {
            for (int j=0; j<board[0].length; j++) {
                if (board[i][j] == 'X') {
                    if (i-1>=0 && board[i-1][j] == 'X') {
                        continue;
                    }
                    if (j-1>=0 && board[i][j-1] == 'X') {
                        continue;
                    }
                    count ++;
                }
            }
        }
        return count;
    }
}