面试题17.14.最小K个数

链接：面试题17.14.最小K个数
难度：Medium
标签：数组、分治、快速选择、排序、堆（优先队列）
简介：设计一个算法，找出数组中最小的 k 个数。以任意顺序返回这 k 个数均可。

题解 1 - typescript

• 编辑时间：2021-05-07
• 执行用时：164ms
• 内存消耗：45.1MB
• 编程语言：typescript
• 解法介绍：利用内部排序。

function smallestK(arr: number[], k: number): number[] {
  return arr.sort((a, b) => a - b).slice(0, k);
}

题解 2 - typescript

• 编辑时间：2021-09-03
• 执行用时：8000ms
• 内存消耗：48.4MB
• 编程语言：typescript
• 解法介绍：堆。

class Heap<T = number> {
  private arr: T[] = [];
  get isEmpty() {
    return this.size === 0;
  }
  get size() {
    return this.arr.length;
  }
  get top() {
    return this.arr[0];
  }
  constructor(private compare: (t1: T, t2: T) => number) {}
  add(num: T): void {
    this.arr.push(num);
    this.shiftUp(this.size - 1);
  }
  remove(): T {
    const num = this.arr.shift()!;
    if (this.size) {
      this.arr.unshift(this.arr.pop()!);
      this.shiftDown(0);
    }
    return num;
  }
  private shiftUp(index: number): void {
    if (index === 0) return;
    const parentIndex = (index - 1) >> 1;
    if (this.compare(this.arr[index], this.arr[parentIndex]) > 0) {
      [this.arr[index], this.arr[parentIndex]] = [this.arr[parentIndex], this.arr[index]];
      this.shiftUp(parentIndex);
    }
  }
  private shiftDown(index: number): void {
    let childrenIndex = index * 2 + 1;
    if (childrenIndex > this.size - 1) return;
    if (
      childrenIndex + 1 <= this.size - 1 &&
      this.compare(this.arr[childrenIndex + 1], this.arr[childrenIndex]) > 0
    ) {
      childrenIndex++;
    }
    if (this.compare(this.arr[childrenIndex], this.arr[index]) > 0) {
      [this.arr[childrenIndex], this.arr[index]] = [this.arr[index], this.arr[childrenIndex]];
      this.shiftDown(childrenIndex);
    }
  }
  *[Symbol.iterator](): IterableIterator<T> {
    for (const t of this.arr) {
      yield t;
    }
  }
}
function smallestK(arr: number[], k: number): number[] {
  const heap = new Heap((t1, t2) => t2 - t1);
  arr.forEach(v => heap.add(v));
  const ans: number[] = [];
  while (k--) ans.push(heap.remove());
  return ans;
}