計算戰艦的個數 Battleships in a Board

問題：

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?

解決：

① 求戰艦的個數，所謂的戰艦就是隻能是一行或者一列，不能有拐彎。本題限定了不會有相鄰的兩個戰艦的存在，有了這一點限制，那麼咱們只須要遍歷一次二維數組就好了，只要找出戰艦的起始點。所謂的戰艦起始點，就是指一條軍艦上最左邊的那個‘X’或者最上面的那個‘X’。

class Solution { //5ms
    public int countBattleships(char[][] board) {
        if (board == null || board.length == 0 || board[0].length == 0) return 0;
        int count = 0;
        int m = board.length;
        int n = board[0].length;
        for (int i = 0;i < m;i ++){
            for (int j = 0;j < n;j ++){
                if (board[i][j] == '.' || (i > 0 && board[i - 1][j] == 'X') || (j > 0 && board[i][j - 1] == 'X')){
                    continue;
                }
                count ++;
            }
        }
        return count;
    }
}

class Solution {//4ms     public int countBattleships(char[][] board) {         if(board.length == 0) return 0;         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') {                     if(i > 0 && board[i - 1][j] == 'X') continue;                     if(j > 0 && board[i][j - 1] == 'X') continue;                     count ++;                 }             }         }         return count;     } }