【leetcode】1311. Get Watched Videos by Your Friends

题目如下：

There are n people, each person has a unique id between 0 and n-1. Given the arrays watchedVideos and friends, where watchedVideos[i] and friends[i] contain the list of watched videos and the list of friends respectively for the person with id = i.

Level 1 of videos are all watched videos by your friends, level 2 of videos are all watched videos by the friends of your friends and so on. In general, the level k of videos are all watched videos by people with the shortest path equal to k with you. Given your id and the level of videos, return the list of videos ordered by their frequencies (increasing). For videos with the same frequency order them alphabetically from least to greatest.  

Example 1:

Input: watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 1
Output: ["B","C"] 
Explanation: 
You have id = 0 (green color in the figure) and your friends are (yellow color in the figure):
Person with id = 1 -> watchedVideos = ["C"] 
Person with id = 2 -> watchedVideos = ["B","C"] 
The frequencies of watchedVideos by your friends are: 
B -> 1 
C -> 2

Example 2:

Input: watchedVideos = [["A","B"],["C"],["B","C"],["D"]], friends = [[1,2],[0,3],[0,3],[1,2]], id = 0, level = 2
Output: ["D"]
Explanation: 
You have id = 0 (green color in the figure) and the only friend of your friends is the person with id = 3 (yellow color in the figure).

Constraints:

• n == watchedVideos.length == friends.length
• 2 <= n <= 100
• 1 <= watchedVideos[i].length <= 100
• 1 <= watchedVideos[i][j].length <= 8
• 0 <= friends[i].length < n
• 0 <= friends[i][j] < n
• 0 <= id < n
• 1 <= level < n
• if friends[i] contains j, then friends[j] contains i

解题思路：先用BFS求出对应level的friends，然后统计这些friends观看的所有videos，最后排序即可。

代码如下：

class Solution(object):
    def watchedVideosByFriends(self, watchedVideos, friends, id, level):
        """
        :type watchedVideos: List[List[str]]
        :type friends: List[List[int]]
        :type id: int
        :type level: int
        :rtype: List[str]
        """
        level_list = [float('inf')] * 101
        dic_video = {}
        queue = [(id,0)]
        level_list[id] = 0

        while len(queue) > 0:
            p_inx,p_level = queue.pop(0)
            for friend in friends[p_inx]:
                if level_list[friend] > p_level + 1:
                    queue.append((friend,p_level+1))
                    level_list[friend] = p_level+1
        for inx in range(len(level_list)):
            if level_list[inx] == level:
                for video in watchedVideos[inx]:
                    dic_video[video] = dic_video.setdefault(video,0) + 1

        res = []

        releation_list = []
        for key,val in dic_video.iteritems():
            releation_list.append((key,val))

        def cmpf(item1,item2):
            if item1[1] != item2[1]:
                return item1[1] - item2[1]
            if item1[0] > item2[0]:
                return 1
            elif item1[0] <item2[0]:
                return -1
            return 0

        releation_list.sort(cmp=cmpf)
        for key,val in releation_list:
            res.append(key)

        return res