347. Top K Frequent Elements 出现频率最高的k个元素

 Given a non-empty array of integers, return the k most frequent elements.

For example,
Given [1,1,1,2,2,3] and k = 2, return [1,2].

Note:

  • You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
  • Your algorithm's time complexity must be better than O(n log n), where n is the array's size.
例如，给定[1,1,1,2,2,3]和k = 2，返回[1,2]。
注意：
你可以假设k总是有效的，1≤k≤唯一元素的个数。
你的算法的时间复杂度必须比O（n log n）好，其中n是数组的大小。
  1. /**
  2.  * @param {number[]} nums
  3.  * @param {number} k
  4.  * @return {number[]}
  5.  */
  7. let left = (i) => 2 * i + 1;
  8. let right = (i) => 2 * (i + 1);
  10. let buildHeap = (x, less_p) => {
  11.     let n = x.length;
  12.     for (let i = Math.floor(n / 2) - 1; i >= 0; i--) {
  13.         heapify(x, i, less_p)
  14.     }
  15. }
  17. let heapify = (x, i, less_p) => {
  18.     let n = x.length;
  19.     while (true) {
  20.         let l = left(i);
  21.         let r = right(i);
  22.         let smallest = i;
  23.         if (l < n && less_p(x[l], x[i])) {
  24.             smallest = l;
  25.         }
  26.         if (r < n && less_p(x[r], x[smallest])) {
  27.             smallest = r;
  28.         }
  29.         if (smallest != i) {
  30.             [x[i], x[smallest]] = [x[smallest], x[i]];
  31.             i = smallest;
  32.         } else {
  33.             break;
  34.         }
  35.     }
  36. }
  38. let heapPop = (x, less_p) => {
  39.     top = x[0]
  40.     x[0] = x[x.length - 1];
  41.     x.pop()
  42.     if (x.length > 0) {
  43.         heapify(x, 0, less_p)
  44.     }
  45.     return top
  46. }
  48. let topK = (x, k, less_p) => {
  49.     let res = [];
  50.     buildHeap(x, less_p);
  51.     let count = x.length;
  52.     for (let i = 0; i < Math.min(k, count); i++) {
  53.         res.push(heapPop(x, less_p));
  54.     }
  55.     return res;
  56. }
  58. var topKFrequent = function (nums, k) {
  59.     let MAX_HEAP = (a, b) => m[a] > m[b];
  61.     let m = {};
  62.     let heap = [];
  63.     for (let i in nums) {
  64.         let c = nums[i];
  65.         !m[c] && heap.push(c);
  66.         m[c] = m[c] ? ++m[c] : 1;
  67.     }
  68.     return topK(heap, k, MAX_HEAP);
  69. };
  71. let nums = [1, 2];
  72. let k = 2;
  73. let res = topKFrequent(nums, k);
  74. console.log(res);







posted @ 2017-12-26 00:14  xiejunzhao  阅读(244)  评论(0)  编辑  收藏  举报