347. Top k high frequency elements

Find the number of elements , Think of using map Recording .
front k High frequency elements ： Sort .
map Cannot sort values , Replaceable map Key position , Or a map Turn into list To achieve .

 class Solution {
        public int[] topKFrequent(int[] nums, int k) {
            Map<Integer,Integer> map = new HashMap();
            // Record times 
            for(int i : nums){
                if(map.containsKey(i)){
                    map.put(i,map.get(i)+1);
                }else{
                    map.put(i,1);
                }
            }

			// Sort 
            List<Map.Entry<Integer, Integer>> list = new ArrayList<Map.Entry<Integer, Integer>>(map.entrySet());
            Collections.sort(list,new Comparator<Map.Entry<Integer,Integer>>() {
                public int compare(Map.Entry<Integer, Integer> o1, Map.Entry<Integer, Integer> o2) {
                    return o2.getValue()-o1.getValue();
                }
            });
            int i =0;
            nums = new int[k];
            for (Map.Entry<Integer, Integer> e: list) {
                nums[i++]=e.getKey();
                if(i>=k){
                    break;
                }
            }
            return nums;
        }
    }

Knowledge point :
map Cannot sort values directly , We need to convert to list, Use Collections.sort Sort .
The sorting time is either greater than , Or less than , Or equal to , There can be no ambiguity , Unable to determine the size .
o1>o2 It's in ascending order .

Other answers ：

The most appropriate data structure ： Priority queue .（ Pile up , Perfect binary tree ）, It can sort the data .

PriorityQueue<Map.Entry<Integer,Integer>> queue =
new PriorityQueue<>((o1,o2)-> o1.getValue-o2.getValue);

Knowledge point ：
Priority queues use
Rewrite sorting method
offer == set
poll == get And delete