[leetcode] 419. Battleships in a Board @ python_419. battleships in a board python-优快云博客

本文介绍了一种算法问题：在一个二维棋盘上计算战舰的数量。提供了两种解决方案：一是通过遍历棋盘并标记相连的战舰；二是仅记录每个战舰的第一个字符。这两种方法都有效地解决了问题，并附带了详细的Python代码实现。

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1

遍历board, 每次遇到’X’时, count递增, 并将它相连的’X’全部标记为’#’.

代码

class Solution(object):
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """
        def helper(x, y):            
            board[x][y] = '#'
            for dx, dy in [(-1,0), (1,0), (0,-1), (0,1)]:
                while 0<=x+dx<row and 0<=y+dy<col and board[x+dx][y+dy] == 'X':
                    x = x+dx
                    y = y+dy
                    board[x][y] = '#'                   
                    
                    
        row, col = len(board), len(board[0])
        count = 0
        for i in range(row):
            for j in range(col):
                if board[i][j] == 'X':
                    # mark the battleship
                    dfs(i, j)
                    count += 1
                    
        return count

解法2

遍历board, 每次只记录第一个遇到的X, 对于它后面相连的X, 无论是水平相连还是垂直相连, 都不计入count.

代码

class Solution(object):
    def countBattleships(self, board):
        """
        :type board: List[List[str]]
        :rtype: int
        """               
                   
        row, col = len(board), len(board[0])
        count = 0
        for i in range(row):
            for j in range(col):
                if board[i][j] == 'X':
                    flag = 1
                    # just count the first char of the row
                    if j >= 1 and board[i][j-1] == 'X': 
                        flag = 0
                    if i >= 1 and board[i-1][j] == 'X':
                        flag = 0
                    count += flag
                    
        return count