1311. Get Watched Videos by Your Friends

问题：

朋友看视频问题。

给定一个朋友关系网，

friends[i]表示 i 的朋友们。

watchedVideos[i]表示 i 所看的视频。

level表示朋友层，1:代表i的朋友，2:代表i的朋友的朋友。

求给定人 i 的朋友层level，所看过的所有视频。（按照视频被看的人数从少到多排序）

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

　　

example 1:

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example 2:

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解法：BFS

状态：当前人 id

visited：保存遍历过的人 id

level按照queue遍历层数进行递减。

当level递减到0，开始记录该层人所看过的视频。

存入unordered_map中，key：视频，value：看过的人数

每层遍历完，如果level==0，退出遍历。

将结果存入set中，进行排序（first：看过的人数。second：视频）

再将set转存入结果vector<string> 即为已排过序的结果。

代码参考：

class Solution {
public:
    vector<string> watchedVideosByFriends(vector<vector<string>>& watchedVideos, vector<vector<int>>& friends, int id, int level) {
        queue<int> q;
        unordered_set<int> visited;
        unordered_map<string,int> res_map;
        vector<string> res;
        q.push(id);
        visited.insert(id);
        while(!q.empty()) {
            int sz = q.size();
            for(int i=0; i<sz; i++) {
                int cur = q.front();
                q.pop();
                //cout<<"pop:"<<cur<<" level:"<<level<<endl;
                if(level==0) {
                    for(auto vd:watchedVideos[cur])
                        res_map[vd]++;
                } else {
                    for(auto frd:friends[cur]) {
                        if(visited.insert(frd).second) {
                            q.push(frd);
                        }
                    }
                }
            }
            if(level==0) break;
            level--;
        }
        set<pair<int,string>> res_set;
        for(auto rm:res_map) {
            res_set.insert({rm.second,rm.first});
        }
        for(auto rs:res_set) {
            res.push_back(rs.second);
        }
        return res;
    }
};