347. 前 K 个高频元素

Posted 2022-03-07 沿着路走到底

力扣

给你一个整数数组 nums 和一个整数 k ，请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。

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]

class MinHeap 
  constructor() 
    this.heap = []
  

  swap(parentIndex, childIndex) 
    const temp = this.heap[parentIndex]
    this.heap[parentIndex] = this.heap[childIndex]
    this.heap[childIndex] = temp
  

  getParentIndex(index) 
    //return Math.floor((index - 1) / 2)
    return (index - 1) >> 1
  

  getLeftIndex(index) 
    return index * 2 + 1
  

  getRightIndex(index) 
    return index * 2 + 2
  

  shiftUp(index) 
    if (index === 0) return

    const parentIndex = this.getParentIndex(index)
    if(this.heap[parentIndex] && this.heap[parentIndex].value > this.heap[index].value) 
      this.swap(parentIndex, index)
      this.shiftUp(parentIndex)
    
  

  shiftDown(index) 
    const leftIndex = this.getLeftIndex(index)
    const rightIndex = this.getRightIndex(index)

    if(this.heap[leftIndex] && this.heap[index].value > this.heap[leftIndex].value) 
      this.swap(index, leftIndex)
      this.shiftDown(leftIndex)
    

    if(this.heap[rightIndex] && this.heap[index].value > this.heap[rightIndex].value) 
      this.swap(index, rightIndex)
      this.shiftDown(rightIndex)
    
  

  insert (value) 
    this.heap.push(value)
    this.shiftUp(this.heap.length - 1)
  

  pop () 
    this.heap[0] = this.heap.pop()
    this.shiftDown(0)
  

  peek() 
    return this.heap[0]
  

  size() 
    return this.heap.length
  
  


/**
 * @param number[] nums
 * @param number k
 * @return number[]
 */
var topKFrequent = function(nums, k) 
  const map = new Map()
  nums.forEach(n => 
    map.set(n, map.has(n) ? map.get(n)+1 : 1)
  )
  
  const heap = new MinHeap()
  map.forEach((value, key) => 
    heap.insert(value, key)
    if (heap.size() > k) 
      heap.pop()
    
  )

  return heap.heap.map(item => item.key)
;

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