347.go

package main

import (
	"fmt"
	"container/heap"
)

func main() {
	fmt.Println(topKFrequent([]int{1, 2, 1, 2, 1, 2, 1, 3}, 2))
}

//给定一个非空的整数数组，返回其中出现频率前 k 高的元素。
//
//示例 1:
//
//输入: nums = [1,1,1,2,2,3], k = 2
//输出: [1,2]
//示例 2:
//
//输入: nums = [1], k = 1
//输出: [1]
//说明：
//
//你可以假设给定的 k 总是合理的，且 1 ≤ k ≤ 数组中不相同的元素的个数。
//你的算法的时间复杂度必须优于 O(n log n) , n 是数组的大小。
//
//
//链接：https://leetcode-cn.com/problems/top-k-frequent-elements
type Item struct {
	value    int
	priority int
	index    int
}

type PriorityQueue []*Item

func (pq PriorityQueue) Len() int { return len(pq) }

func (pq PriorityQueue) Less(i, j int) bool {
	return pq[i].priority > pq[j].priority
}

func (pq PriorityQueue) Swap(i, j int) {
	pq[i], pq[j] = pq[j], pq[i]
	pq[i].index = i
	pq[j].index = j
}

func (pq *PriorityQueue) Push(x interface{}) {
	n := len(*pq)
	item := x.(*Item)
	item.index = n
	*pq = append(*pq, item)
}

func (pq *PriorityQueue) Pop() interface{} {
	old := *pq
	n := len(old)
	item := old[n-1]
	item.index = -1
	*pq = old[0 : n-1]
	return item
}

func (pq *PriorityQueue) update(item *Item, value int, priority int) {
	item.value = value
	item.priority = priority
	heap.Fix(pq, item.index)
}

func topKFrequent(nums []int, k int) []int {
	timesMap := make(map[int]int, 0)
	for _, num := range nums {
		timesMap[num]++
	}
	pq := make(PriorityQueue, len(timesMap))
	i := 0
	for value, priority := range timesMap {
		pq[i] = &Item{
			value:    value,
			priority: priority,
			index:    i,
		}
		i++
	}
	heap.Init(&pq)
	result := make([]int, 0)
	for k > 0 {
		result = append(result, heap.Pop(&pq).(*Item).value)
		k--
	}
	return result
}