leetcode 419. Battleships in a Board

Question

Given an 2D board, count how many different 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.

Could you do it in one-pass, using only O(1) extra memory and without modifying the value of the board?

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.

问题

现在用一个包含’X’和’.’的二维字符矩阵来表示一组战舰群。每条战舰用横向或纵向的’X’表示。每条战舰之间至少用一个’.’字符隔开。问是否能够在一次遍历、O(1)空间复杂度、不修改输入的情况下找出有多少条战舰。

分析

我首先思考的问题是为什么要用一个‘.’将战舰隔开。原因很简单，因为如果战舰相互邻接，战舰的数量将会变得不确定。
那么如何才能在遍历一次情况下统计出战舰的数量呢？有一个很巧妙的想法：数船头。因为船头是非常好判断的，船头的左边和上边必然是’.’，而且这是个充要条件。船身必然不符合这个条件，因为横向战舰的船身的左边一定是’X’，纵向战舰的船身的上边一定是’X’。这样一来，只要一次遍历整个二维数组，统计船头的个数就行了。

代码

int countBattleships(vector<vector<char>>& board) {        int cnt = 0;        for (int i = 0; i < board.size(); i++)        {            vector<char> &row = board[i];            for (int j = 0; j < row.size(); j++)            {                if (board[i][j] == 'X' &&                (i == 0 || (i > 0 && board[i - 1][j] == '.')) &&                (j == 0 || (j > 0 && board[i][j - 1] == '.')))                {                    cnt++;                }            }        }        return cnt;    }

总结

如果要判断某一复杂物体的类型，只要判断它最有特征性、最容易判断的一部分就行了（似乎很有哲理的样子呢>。<）