419. Battleships in a Board

标签：return   boa   using   call   without   style   解题思路   +=   extra   

问题描述：

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?

解题思路：

因为题目中告诉我们，给我们的是一个有效的board，也就是说，两艘战舰之间必然有空隙。

我们可以将横着的最左作为开头，竖着的最上作为开头

我们对每一个‘X’检查它是否为开头，若是则+1.

代码：

class Solution {
public:
    int countBattleships(vector<vector<char>>& board) {
        int m = board.size();
        int n = board[0].size();
        int ret = 0;
        for(int i = 0; i < m; i++){
            for(int j = 0; j < n; j++){
                ret += board[i][j] == ‘X‘ && (i == 0 || board[i-1][j] != ‘X‘) && (j == 0 || board[i][j-1] != ‘X‘);
            }
        }
        return ret;
    }
};

419. Battleships in a Board

标签：return   boa   using   call   without   style   解题思路   +=   extra   

原文地址：https://www.cnblogs.com/yaoyudadudu/p/9227127.html