LeetCode 347. Top K Frequent Elements（最频繁的K个元素）

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

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.

使用两个映射：num --> frequency，和frequency --> nums

public class Solution {    public List<Integer> topKFrequent(int[] nums, int k) {        Map<Integer, Integer> frequency = new HashMap<>();        int max = 0;        for(int num: nums) {            Integer f = frequency.get(num);            if (f == null) f = 1; else f ++;            frequency.put(num, f);            max = Math.max(max, f);        }        List<Integer>[] inverse = new List[max+1];        for(int num: frequency.keySet()) {            int f = frequency.get(num);            if (inverse[f] == null) inverse[f] = new ArrayList<>();            inverse[f].add(num);        }        List<Integer> result = new ArrayList<>();        for(int i=max; i>0; i--) {            if (result.size() == k) break;            if (inverse[i] == null) continue;            for(int j=0; j<inverse[i].size(); j++) {                if (result.size() == k) break;                result.add(inverse[i].get(j));            }        }        return result;    }}

方法二：对哈希映射计数表进行排序。

public class Solution {    public List<Integer> topKFrequent(int[] nums, int k) {        Map<Integer, Integer> frequency = new HashMap<>();        for(int num: nums) {            Integer f = frequency.get(num);            if (f == null) f = 1; else f ++;            frequency.put(num, f);        }        int[] uniques = new int[frequency.size()];        int[] f = new int[uniques.length];        int i = 0;        for(Map.Entry<Integer, Integer> entry: frequency.entrySet()) {            uniques[i] = entry.getKey();            f[i] = entry.getValue();            i++;        }        Integer[] n = new Integer[uniques.length];        for(int j=0; j<n.length; j++) n[j] = j;        Arrays.sort(n, new Comparator<Integer>() {            @Override            public int compare(Integer i1, Integer i2) {                return Integer.compare(f[i2], f[i1]);            }        });        List<Integer> results = new ArrayList<>();        for(int j=0; j<k && j<n.length; j++) {            results.add(uniques[n[j]]);        }        return results;    }}

不合适使用摩尔投票算法，因为题目是找最频繁，而不是找频率超过1/k的数字。