LeetCode——419. Battleships in a Board

题型：DFS

题目：

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?

 

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翻译（抄的）：

给定一个2维板，计算其中包含多少艘不一样的战舰。战舰用字符'X'表示，空白槽位用'.'表示。你应该假设以下规则：

• 给定的板子是有效的，只包含战舰和空白槽位。
• 战舰只能水平或者竖直放置。
• 战舰的尺寸可能不一样。
• 两艘战舰之间在水平方向或者竖直方向至少包含一个空间—不会存在相邻的战舰。

你的算法不该该修改board的值。

 

思路：

一种思路是简单遍历，由于题目给了方便，战舰之间没有相邻的（这里要理解，战舰是1-N个X组成，而不是一个）

 

另外一种思路是DFS深搜，归纳来说就是对每个点去深搜遍历它全部的相邻点，经过一个set来维护遍历过的点，这样每个点能够获得它相邻的全部点，即一艘战舰，再检查下一个点就能够先判断在不在set里，是否是已经属于统计过的某个战舰了。

 

代码（python）：

class Solution(object):
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        found_set = set()
        ans = 0
        row_len = len(board)
        col_len = len(board[0] if row_len else [])
        def dfs(x, y):
            for movex, movey in zip((1,0,-1,0),(0,1,0,-1)):
                nx, ny = x + movex, y + movey
                if 0 <= nx < row_len and 0 <= ny < col_len:
                    if (nx,ny) not in found_set and board[nx][ny] == 'X':
                        found_set.add((nx, ny))
                        dfs(nx,ny)
                        
        for x in xrange(row_len):
            for y in xrange(col_len):
                if board[x][y] == 'X' and (x,y) not in found_set:
                    ans += 1
                    found_set.add((x,y))
                    dfs(x,y)
        return ans

 

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