419. Battleships in a Board 棋盘上的战舰数量

［抄题］：

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.

[暴力解法]：

时间分析：

空间分析：

[优化后]：

时间分析：

空间分析：

［奇葩输出条件］：

［奇葩corner case］：

i - 1 时请务必做好检查也要>=0，就是i >= 1

if (i >= 1 && board[i - 1][j] == 'X') continue;
if (j >= 1 && board[i][j - 1] == 'X') continue;

［思维问题］：

以为要用dfs，但是题目对所在的位置有特殊要求，就只能老老实实一个个地数了

［英文数据结构或算法，为什么不用别的数据结构或算法］：

［一句话思路］：

同一行或者同一列，就只数头不数尾

［输入量］：空： 正常情况：特大：特小：程序里处理到的特殊情况：异常情况（不合法不合理的输入）：

［画图］：

［一刷］：

［二刷］：

［三刷］：

［四刷］：

［五刷］：

[五分钟肉眼debug的结果]：

［总结］：

题目对所在的位置有特殊要求，就只能老老实实一个个地数了

［复杂度］：Time complexity: O(mn) Space complexity: O(1)

［算法思想：迭代/递归/分治/贪心］：

［关键模板化代码］：

［其他解法］：

［Follow Up］：

［LC给出的题目变变变］：

[代码风格] ：

[是否头一次写此类driver funcion的代码] ：

[潜台词] ：

class Solution {
    public int countBattleships(char[][] board) {
        //ini some variables
        int m = board.length; int n = board[0].length;
        int count = 0;

        //cc
        if (board == null || m == 0 || n == 0) return 0; 

        //for loop, only add once if qualified
        for(int i = 0; i < m; i++) {
            for (int j = 0; j < n; j++) {
                if (board[i][j] == '.') continue;
                if (i >= 1 && board[i - 1][j] == 'X') continue;
                if (j >= 1 && board[i][j - 1] == 'X') continue;

                //add
                count++;
            }
        }

        //return
        return count;
    }
}