How to find friends

How to find friends

 

Simple thinking and hard Coding

 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 True22 23         for each in extend:24             if each not in visited:25                 future.append(each)26 27     return False

Use dict to store the direct relationship of each person, such as {lyly: [Lala, Gege]}. Use two lists, one of which indicates the person who has already traversed, and the other that is going to traverse;

In addition, two-dimensional arrays in python can be defined as follows: Connection = [[false for col in range (5)] For row in range (5)], which is a 5*5 array, I tried to define error = [[false] * 5] * 5 in this way. I found that I only needed to modify it.

Error [0] [0], then error [1] [0], error [2] [0]... because [false] * 5 generates a one-dimensional array, and * 5 generates only five references. Therefore, no matter which one is modified, the remaining one changes.

 

Measure the test taker's knowledge about the code.

 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, b11 12     return any(set([first, second]) <= s for s in setlist)

Use each link as a set. If the intersection between links is not empty, calculate the Union of links, that is, the link circle;

Finally, check whether first and second are in a link circle;

Make full use of the set operation, & intersection, | and, <= subset;

There are also 12th rows of syntax features