[leetcode] 419. Battleships in a Board @ python

最新推荐文章于 2024-07-24 17:53:27 发布

原创最新推荐文章于 2024-07-24 17:53:27 发布 · 196 阅读

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本文介绍了一种使用深度优先搜索(DFS)算法在一维数组表示的二维棋盘上计数战舰数量的方法。战舰由'X'表示，空位由'.'表示。文章详细解释了算法原理，即遍历棋盘，遇到'X'时增加计数，并标记其相邻的'X'，避免重复计数。最终返回计数值。

原题

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?

解法

DFS. 遍历board, 遇到值为’X’的格子时, res加一, 并且将它相邻的’X’标记为’#’. DFS的终止条件是, 当i 或者j 超出index范围或者board[i][j] 不为’X’时, return.
Time: O(row * col)
Space: O(1)

代码

class Solution:
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        def dfs(i, j):
            if 0 <= i < row and 0 <= j < col and board[i][j] == 'X':
                board[i][j] = '#'
                dfs(i-1, j)
                dfs(i+1, j)
                dfs(i, j+1)
                dfs(i, j-1)
            else:
                return
                
        row, col = len(board), len(board[0])
        res = 0
        for i in range(row):
            for j in range(col):
                if board[i][j] == 'X':
                    res += 1
                    # mark the horizontally or vertically X as readed
                    dfs(i, j)
                    
        return res