[路飞]leetcode-373.查找和最小的K对数字

给定两个以 升序排列 的整数数组 nums1 和 ****nums2 ****, 以及一个整数 k ****。

定义一对值 (u,v)，其中第一个元素来自 nums1，第二个元素来自 nums2 ****。

请找到和最小的 k 个数对 (u1,v1),  (u2,v2)  ...  (uk,vk) 。 力扣原文

示例 1:

输入: nums1 = [1,7,11], nums2 = [2,4,6], k = 3
输出: [1,2],[1,4],[1,6]
解释: 返回序列中的前 3 对数：
     [1,2],[1,4],[1,6],[7,2],[7,4],[11,2],[7,6],[11,4],[11,6]

示例 2:

输入: nums1 = [1,1,2], nums2 = [1,2,3], k = 2
输出: [1,1],[1,1]
解释: 返回序列中的前 2 对数：
     [1,1],[1,1],[1,2],[2,1],[1,2],[2,2],[1,3],[1,3],[2,3]

示例 3:

输入: nums1 = [1,2], nums2 = [3], k = 3 
输出: [1,3],[2,3]
解释: 也可能序列中所有的数对都被返回:[1,3],[2,3]

解题利用顶堆:

var kSmallestPairs = function (nums1, nums2, k) {
  let heap = new BigHeap(k);
  for (let i = 0; i < nums1.length; i++) {
    for (let j = 0; j < nums2.length; j++) {
      const item = [nums1[i], nums2[j]];
      if (heap.size < k || compare(heap.top(), item)) {
        heap.push(item);
      } else {
        break;
      }
    }
  }
  return heap.arr;
};
function compare(arr1, arr2) {
  return arr1[0] + arr1[1] > arr2[0] + arr2[1];
}
class BigHeap {
  constructor(k) {
    this.arr = [];
    this.max = k;
    this.size = 0;
  }
  top() {
    return this.arr[0];
  }
  push(val) {
    this.arr.push(val);
    this.size++;
    if (this.size > 1) {
      let cur = this.size - 1,
      pre = (cur - 1) >> 1;
      while (cur > 0 && compare(this.arr[cur], this.arr[pre])) {
        [this.arr[cur], this.arr[pre]] = [this.arr[pre], this.arr[cur]];
        cur = pre;
        pre = (cur - 1) >> 1;
      }
    }
    if (this.size > this.max) this.pop();
  }
  pop() {
    this.arr[0] = this.arr.pop();
    this.size--;
    let cur = 0,
      left = cur * 2 + 1,
      right = cur * 2 + 2;
    while (
      (left < this.size && compare(this.arr[left], this.arr[cur])) ||
      (right < this.size && compare(this.arr[right], this.arr[cur]))
    ) {
      if (right < this.size && compare(this.arr[right], this.arr[left])) {
        [this.arr[cur], this.arr[right]] = [this.arr[right], this.arr[cur]];
        cur = right;
      } else {
        [this.arr[cur], this.arr[left]] = [this.arr[left], this.arr[cur]];
        cur = left;
      }
      (left = cur * 2 + 1), (right = cur * 2 + 2);
    }
  }
}