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

How to find friends