LeetCode 刷题笔记之频率最高的k个数（Top K Frequent Elements）

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题目如下：

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.

解答如下：

复制代码

class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        HashMap<Integer,Integer> Freq=new HashMap<Integer,Integer>();//构建频率表
        for(int i=0;i<nums.length;i++){
            if(Freq.containsKey(nums[i])){
                int tmp=Freq.get(nums[i]);
                Freq.remove(nums[i]);
                Freq.put(nums[i],tmp+1);//频率加一
            }
            else Freq.put(nums[i],1);
        }
        List<Map.Entry<Integer,Integer>> list = new ArrayList<Map.Entry<Integer,Integer>>(Freq.entrySet());
        Collections.sort(list,new Comparator<Map.Entry<Integer,Integer>>() {//对map进行排序
            public int compare(Map.Entry<Integer, Integer> o1,
                    Map.Entry<Integer, Integer> o2) {
                return o2.getValue().compareTo(o1.getValue());
            }
        });
        int i=0;
        List<Integer> topk=new ArrayList<Integer>();
        for(Map.Entry<Integer,Integer> mapping:list){ //输出频率最高的k个数
               if(i==k) break;
               topk.add(mapping.getKey()); 
               i++;
        }
        return topk;
    }
}

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class Solution {
    public List<Integer> topKFrequent(int[] nums, int k) {
        HashMap<Integer,Integer> Freq=new HashMap<Integer,Integer>();//构建频率表
        for(int i=0;i<nums.length;i++){
            if(Freq.containsKey(nums[i])){
                int tmp=Freq.get(nums[i]);
                Freq.remove(nums[i]);
                Freq.put(nums[i],tmp+1);//频率加一
            }
            else Freq.put(nums[i],1);
        }
        List<Map.Entry<Integer,Integer>> list = new ArrayList<Map.Entry<Integer,Integer>>(Freq.entrySet());
        Collections.sort(list,new Comparator<Map.Entry<Integer,Integer>>() {//对map进行排序
            public int compare(Map.Entry<Integer, Integer> o1,
                    Map.Entry<Integer, Integer> o2) {
                return o2.getValue().compareTo(o1.getValue());
            }
        });
        int i=0;
        List<Integer> topk=new ArrayList<Integer>();
        for(Map.Entry<Integer,Integer> mapping:list){ //输出频率最高的k个数
               if(i==k) break;
               topk.add(mapping.getKey()); 
               i++;
        }
        return topk;
    }
}