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[video] = dict[video] + 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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