[leetcode]419. Battleships in a Board

Question: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 shape1xN (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:

...XXXXX...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 andwithout modifying the value of the board?

Analysis:这个题，就是用矩阵一个放战舰的地方，X是放的地方，.是空槽

他肯定给成横着的一条或者竖着的一条，而且肯定有隔离，不会交叉。。现在求出有多少个战舰

因为给定的数据肯定是有效的，所以我们直接统计最左上的节点的数量就好，就是每一个战舰的最最上角，统计一下数额就是了

Answer:

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++) {                //找到最top-left的那一个，并且判断其是否和合法，也就是没有交叉                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;    }}