[LeetCode] Top K Frequent Elements

来源：互联网发布：大数据分析师待遇编辑：程序博客网时间：2024/05/19 19:39

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.

解题思路

用一个Map统计每个数字的出现频率。
依据出现频率对Map.Entry进行排序，排序算法为堆排序。
输出出现频率最高的k个元素。

实现代码

//Runtime: 43 mspublic class Solution {    private void buildHeap(ArrayList<Map.Entry<Integer, Integer>> entries, int root, int len) {        int left = 2 * root + 1;        if (left < len) {            int largest = left;            int right = left + 1;            if (right < len && entries.get(right).getValue() > entries.get(largest).getValue()) {                largest = right;            }            if (entries.get(root).getValue() < entries.get(largest).getValue()) {                Map.Entry<Integer, Integer> entry = entries.get(root);                entries.set(root, entries.get(largest));                entries.set(largest, entry);                buildHeap(entries, largest, len);            }        }    }    private void heapSort(ArrayList<Map.Entry<Integer, Integer>> entries) {        int len = entries.size();        for (int i = len / 2; i >= 0; i--) {            buildHeap(entries, i, len);        }        for (int i = len - 1; i > 0; i--) {            Map.Entry<Integer, Integer> entry = entries.get(0);            entries.set(0, entries.get(i));            entries.set(i, entry);            buildHeap(entries, 0, --len);        }    }    public List<Integer> topKFrequent(int[] nums, int k) {        Map<Integer, Integer> map = new HashMap<Integer, Integer>();        for (int num : nums) {            if (map.containsKey(num)) {                map.put(num, map.get(num) + 1);            } else {                map.put(num, 1);            }        }        ArrayList<Map.Entry<Integer, Integer>> entries = new ArrayList<Map.Entry<Integer, Integer>>(map.entrySet());        heapSort(entries);        List<Integer> list = new ArrayList<Integer>(k);        for (int i = entries.size() - 1; i >= entries.size() - k; i--) {            list.add(entries.get(i).getKey());        }        return list;    }}

0 0