From 47c4cc102123b1b43f09b7b766fdd2f1e1b979a9 Mon Sep 17 00:00:00 2001 From: ZongqinWang <1722249371@qq.com> Date: Tue, 17 May 2022 19:52:32 +0800 Subject: [PATCH] =?UTF-8?q?=E6=B7=BB=E5=8A=A0=200347.=E5=89=8DK=E4=B8=AA?= =?UTF-8?q?=E9=AB=98=E9=A2=91=E5=85=83=E7=B4=A0.md=20Scala=E7=89=88?= =?UTF-8?q?=E6=9C=AC?= MIME-Version: 1.0 Content-Type: text/plain; charset=UTF-8 Content-Transfer-Encoding: 8bit --- problems/0347.前K个高频元素.md | 42 ++++++++++++++++++++++++++++ 1 file changed, 42 insertions(+) diff --git a/problems/0347.前K个高频元素.md b/problems/0347.前K个高频元素.md index 1d6a358b..20932b28 100644 --- a/problems/0347.前K个高频元素.md +++ b/problems/0347.前K个高频元素.md @@ -374,7 +374,49 @@ function topKFrequent(nums: number[], k: number): number[] { }; ``` +Scala: +解法一: 优先级队列 +```scala +object Solution { + import scala.collection.mutable + def topKFrequent(nums: Array[Int], k: Int): Array[Int] = { + val map = mutable.HashMap[Int, Int]() + // 将所有元素都放入Map + for (num <- nums) { + map.put(num, map.getOrElse(num, 0) + 1) + } + // 声明一个优先级队列，在函数柯里化那块需要指明排序方式 + var queue = mutable.PriorityQueue[(Int, Int)]()(Ordering.fromLessThan((x, y) => x._2 > y._2)) + // 将map里面的元素送入优先级队列 + for (elem <- map) { + queue.enqueue(elem) + if(queue.size > k){ + queue.dequeue // 如果队列元素大于k个，出队 + } + } + // 最终只需要key的Array形式就可以了，return关键字可以省略 + queue.map(_._1).toArray + } +} +``` +解法二: 相当于一个wordCount程序 +```scala +object Solution { + def topKFrequent(nums: Array[Int], k: Int): Array[Int] = { + // 首先将数据变为(x,1)，然后按照x分组，再使用map进行转换(x,sum)，变换为Array + // 再使用sort针对sum进行排序，最后take前k个，再把数据变为x,y,z这种格式 + nums.map((_, 1)).groupBy(_._1) + .map { + case (x, arr) => (x, arr.map(_._2).sum) + } + .toArray + .sortWith(_._2 > _._2) + .take(k) + .map(_._1) + } +} +``` -----------------------