LeetCode 419. Battleships in a Board 题解（C++）

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题目描述

• 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.

举例

• Example:
  X..X
  …X
  …X
  In the above board there are 2 battleships.
• Invalid Example:
  …X
  XXXX
  …X
  This is not a valid board - as battleships will always have a cell separating between them.

思路

• 这里需要计算战舰队的数量count，可转换为计算每个战舰队位于最左上角的船只（我们称为战舰队头）。
• 用两个循环，对网格内的每一个格子进行遍历，若该格子为X（即有战舰），则分以下几种情况：
  若在网格的最左上角（即i=0，j=0），则存在一个舰队，count自加1；
  若在网格的最上角（即i=0），但不在最左边（即j!=0），则需要判断该位置的左边（即board[i][j-1]）是否有战舰存在，若其左边不存在战舰，则该位置为战舰队头，count自加1；
  若在网格的最左边（即j=0），但不在最上边（即i!=0），则需要判断该位置的上面（即board[i-1][j]）是否有战舰存在，若其上面不存在战舰，则该位置为战舰队头，count自加1；
  若战舰不在最左边也不在最上边（即i!=0且j!=0），则需要判断该战舰的左边和上面是否还有战舰存在，若其左边和上面都不存在战舰，则该位置为战舰队头，count自加1；
  其余情况，战舰都不属于战舰队头。

代码

class Solution {public:    int countBattleships(vector<vector<char>>& board)    {        int count = 0;        int row = board.size();        int column = board[0].size();        for (int i = 0; i < row; ++i)        {            for (int j = 0; j < column; ++j)            {                if (board[i][j] == 'X')                {                    if (i == 0 && j == 0)                    {                        ++count;                    }                    else if (i == 0 && board[i][j - 1] != 'X')                    {                        ++count;                    }                    else if (j == 0 && board[i - 1][j] != 'X')                    {                        ++count;                    }                    else if (i != 0 && j != 0 && board[i][j - 1] != 'X' && board[i - 1][j] != 'X')                    {                        ++count;                    }                }            }        }        return count;    }};