leetcode1267. Count Servers that Communicate

Posted 2020-11-22 seyjs

题目如下：

You are given a map of a server center, represented as a m * n integer matrix grid, where 1 means that on that cell there is a server and 0 means that it is no server. Two servers are said to communicate if they are on the same row or on the same column.

Return the number of servers that communicate with any other server. 

Example 1:

Input: grid = [[1,0],[0,1]]
Output: 0
Explanation: No servers can communicate with others.

Example 2:

Input: grid = [[1,0],[1,1]]
Output: 3
Explanation: All three servers can communicate with at least one other server.

Example 3:

Input: grid = [[1,1,0,0],[0,0,1,0],[0,0,1,0],[0,0,0,1]]
Output: 4
Explanation: The two servers in the first row can communicate with each other. The two servers in the third column can communicate
 with each other. The server at right bottom corner can‘t communicate with any other server. 

Constraints:

• m == grid.length
• n == grid[i].length
• 1 <= m <= 250
• 1 <= n <= 250
• grid[i][j] == 0 or 1

解题思路：和求岛的个数，最大岛等题目的解法一样，DFS或者BFS。

代码如下：

class Solution(object):
    def countServers(self, grid):
        """
        :type grid: List[List[int]]
        :rtype: int
        """
        res = 0
        visit = [[0] * len(grid[0]) for _ in grid]
        for i in range(len(grid)):
            for j in range(len(grid[i])):
                if grid[i][j] == 0 or visit[i][j] == 1:
                    continue
                queue = [(i,j)]
                visit[i][j] = 1
                group = 0
                while len(queue) > 0:
                    x,y = queue.pop(0)
                    group += 1
                    for k in range(len(grid[x])):
                        if grid[x][k] == 1 and visit[x][k] == 0:
                            queue.append((x,k))
                            visit[x][k] = 1
                    for k in range(len(grid)):
                        if grid[k][y] == 1 and visit[k][y] == 0:
                            queue.append((k,y))
                            visit[k][y] = 1
                if group > 1:res += group
        return res