How to find friends

How to find friends

 

思路简单,编码不易

 1 def check_connection(network, first, second):

 2     link_dictionary = dict() 3 4 for link in network: 5 drones = link.split('-') 6 7 link_dictionary.setdefault(drones[0], []).append(drones[1]) 8 link_dictionary.setdefault(drones[1], []).append(drones[0]) 9 10 future = [] 11 visited = [] 12  future.append(first) 13 14 while future: 15 current = future.pop() 16  visited.append(current) 17 18 extend = link_dictionary[current] 19 20 if second in extend: 21 return True 22 23 for each in extend: 24 if each not in visited: 25  future.append(each) 26 27 return False

使用dict存储每个人的直接关系, 如{lyly : [lala, gege]}; 使用两个list, 其中一个表示已遍历过的人, 另一个表示即将遍历的人;

另外python中的二维数组可以这么定义: connection = [[False for col in range(5)] for row in range(5)], 即一个5*5的数组, 原先尝试过这么定义error = [[False] * 5] * 5, 发现只要修改了

error[0][0], 那么error[1][0], error[2][0] ...都修改了, 因为[False] * 5产生了一个一维数组, 而外面的*5只是产生了5个引用因此, 无论修改哪一个其余的均会改变.

 

观摩下大神的代码

 1 def check_connection(network, first, second):

 2     setlist = [] 3 for connection in network: 4 s = ab = set(connection.split('-')) 5 # unify all set related to a, b 6 for t in setlist[:]: # we need to use copy 7 if t & ab: # check t include a, b 8 s |= t 9  setlist.remove(t) 10 setlist.append(s) # only s include a, b 11 12 return any(set([first, second]) <= s for s in setlist)

将每条关系作为set, 若关系间交集不为空, 则求关系间的并集, 即关系圈;

最后查看first与second是否在某个关系圈中;

充分利用了set的操作, &交, |并, <=子集;

还有第12行的语法特性