链接: https://leetcode-cn.com/problems/get-watched-videos-by-your-friends/
难度：medium

解题思路：广搜找到对应level的所有朋友，然后累加相应的video，最后排序。go语言没有啥priority queue，图简单这里排序用的直接插入排序，有点挫

Golang的数据结构支持太少了，各种都要自己写，真是麻烦。。

func watchedVideosByFriends(watchedVideos [][]string, friends [][]int, id int, level int) []string {    visited := map[int]int{}    queue := []Person {Person{id, 0}}    visited[id] = 1    for len(queue) > 0 {        friend := queue[0]        if friend.level == level {            break        }        queue = queue[1:]        for _, f := range friends[friend.id] {            if visited[f] == 0 {                queue = append(queue, Person{f, friend.level + 1})                visited[f] = 1            }        }        visited[friend.id] = 1    }    visited = map[int]int{}    dict := map[string]int {}    for _, p := range queue {        if visited[p.id] == 1 {            continue        }        for _, video := range watchedVideos[p.id] {            dict = dict + 1        }        visited[p.id] = 1    }    result := []string {}    for k, v := range dict {        inserted := false        for idx, e := range result {            if v < dict[e] || (v == dict[e] && k < e) {                result = append(result, "")                copy(result[idx + 1:], result[idx:])                result[idx] = k                inserted = true                break            }        }        if !inserted {            result = append(result, k)        }    }    return result}type Person struct {    id int    level int}

执行用时 : 1276 ms , 在所有 Go 提交中击败了 7.14% 的用户
内存消耗 : 6.6 MB , 在所有 Go 提交中击败了 100.00% 的用户

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