leetcode1311

class Solution:
    def getLevelFriends(self,friends,visited,id,level,l):
        while len(l) > 0 and level > 0:
            temp = []
            while len(l) > 0:
                cur = l.pop()
                flist = friends[cur]
                for f in flist:
                    if visited[f] == 0:
                        visited[f] = 1
                        temp.append(f)
            l = temp[:]
            temp.clear()
            level -= 1
        return l


    def watchedVideosByFriends(self, watchedVideos: 'List[List[str]]', friends: 'List[List[int]]', id: int, level: int) -> 'List[str]':
        n = len(watchedVideos)
        visited = [0] * n
        visited[id] = 1
        friendlist = self.getLevelFriends(friends,visited,id,level,[id])
        #print(friendlist)
        videscounts = {}
        for f in friendlist:
            videos = watchedVideos[f]
            for v in videos:
                if v not in videscounts:
                    videscounts[v] = 1
                else:
                    videscounts[v] += 1
        
        result = [v[0] for v in sorted(videscounts.items(),key=lambda x:(x[1],x[0]))] 
        return result

算法思路：BFS + 多条件排序

先使用BFS计算n度人脉，通过visited数组记录已经包含在好友关系中的人，得到n度人脉列表。

再将对列转化成以vidio为key，观看频率为value的字典，然后对字典进行双条件排序，

先按照value从小到大排序，再按照key从小到大排序，最终得到key的列表即为所求。