原文地址： 

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:

...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 and without modifying the value of the board?

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给定一个二维的甲板， 请计算其中有多少艘战舰。 战舰用 'X'表示，空位用 '.'表示。 你需要遵守以下规则：

• 给你一个有效的甲板，仅由战舰或者空位组成。
• 战舰只能水平或者垂直放置。换句话说,战舰只能由 1xN (1 行, N 列)组成，或者 Nx1 (N 行, 1 列)组成，其中N可以是任意大小。
• 两艘战舰之间至少有一个水平或垂直的空位分隔 - 即没有相邻的战舰。

示例 :

X..X...X...X

在上面的甲板中有2艘战舰。

无效样例 :

...XXXXX...X

你不会收到这样的无效甲板 - 因为战舰之间至少会有一个空位将它们分开。

进阶:

你可以用一次扫描算法，只使用O(1)额外空间，并且不修改甲板的值来解决这个问题吗？

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20ms

1 class Solution { 2     func countBattleships(_ board: [[Character]]) -> Int { 3          4         var battle = 0 5          6         for r in 0..
     
       0 {14                         cellLeft = String(board[r][c-1])15                     }16                     if r > 0{17                         cellUp = String(board[r-1][c])18                     }19                     20                     if cellUp == "." && cellLeft == "."{21                         battle += 122                     }23                 }24                 25             }26         }27         28         return battle29     }30 }
     

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28ms

1 class Solution { 2     func countBattleships(_ board: [[Character]]) -> Int { 3         guard !board.isEmpty && !board[0].isEmpty else { 4             return 0 5         } 6  7         var board = board 8         let xChar = Character("X") 9         let dotChar = Character(".")10         var result = 011         for row in 0..

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152ms

1 class Solution { 2     func countBattleships(_ board: [[Character]]) -> Int { 3         let row = board.count 4         guard row > 0 else { 5             return 0 6         } 7         let column = board[0].count 8         var num = 0 9         10         for i in 0..

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172ms

1 class Solution { 2     func countBattleships(_ board: [[Character]]) -> Int { 3         var board = board 4         if board.isEmpty || board[0].isEmpty {
   return 0} 5         var m:Int = board.count 6         var n:Int = board[0].count 7         var res:Int = 0 8         var visited:[[Bool]] = [[Bool]](repeating:[Bool](repeating:false,count:n),count:m) 9         for i in 0..
     
      = m || j < 0 || j >= n || visited[i][j] || board[i][j] == "."33         {34             return35         }36         vertical |= i37         horizontal |= j38         visited[i][j] = true39         dfs(&board, &visited, &vertical, &horizontal, i - 1, j)40         dfs(&board, &visited, &vertical, &horizontal, i + 1, j)41         dfs(&board, &visited, &vertical, &horizontal, i, j - 1)42         dfs(&board, &visited, &vertical, &horizontal, i, j + 1)43     }44 }