leetcode 347. Top K Frequent Elements | 347. 前 K 个高频元素（大根堆）

题目

https://leetcode.com/problems/top-k-frequent-elements/

题解

参考：leetcode 215. Kth Largest Element in an Array | 215. 数组中的第K个最大元素（Java）

class Solution {public static class HeapNode {int num;int count;public HeapNode(int num, Integer count) {this.num = num;this.count = count;}}public int[] topKFrequent(int[] nums, int k) {HashMap<Integer, Integer> map = new HashMap<>();for (int n : nums) {if (!map.containsKey(n)) map.put(n, 1);else map.put(n, map.get(n) + 1);}// 建大根堆int heapSize = map.size();HeapNode[] heap = new HeapNode[heapSize];int j = 0;for (int key : map.keySet())heap[j++] = new HeapNode(key, map.get(key));for (int i = heapSize - 1; i >= 0; i--) // 自下向上建堆 可证复杂度为O(n)heapify(heap, i, heapSize);// Top kint[] result = new int[k];int pos = 0;for (int i = 0; i < k; i++) {result[pos++] = heap[0].num;swap(heap, 0, --heapSize);heapify(heap, 0, heapSize);}return result;}public void heapify(HeapNode[] heap, int i, int heapSize) {int left = i * 2 + 1;while (left < heapSize) {int largest = left + 1 < heapSize && heap[left + 1].count > heap[left].count ? left + 1 : left; // 左右孩子较大者的下标largest = heap[largest].count > heap[i].count ? largest : i; // 两个孩子与父节点较大者的下标if (largest == i) break; // 不需要交换的情况swap(heap, largest, i);i = largest; // 更新i使其下沉left = 2 * i + 1;}}private void swap(HeapNode[] heap, int i, int j) {HeapNode tmp = heap[i];heap[i] = heap[j];heap[j] = tmp;}
}

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leetcode 347. Top K Frequent Elements | 347. 前 K 个高频元素（大根堆）

题目

题解

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