(leetcode347):
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]
.
思路:(常规)遍历数组,记录数值及对应的频率,再排序找出前k个。
用map存储数值及对应频率,用优先队列找出前k个,优先队列可以直接存储map、排序,方便。Java这里的优先队列默认是最小堆实现, 比如对int就是优先级:从小到大。所以自定义比较器时要注意。
public class L347TopKFrequentElements { public static List<Integer> topKFrequent(int[] nums, int k) { List<Integer> list = new ArrayList<Integer>(); HashMap<Integer,Integer> map = new HashMap<Integer,Integer>(); for (int i = 0; i < nums.length; i++) { if(map.containsKey(nums[i])){ map.put(nums[i], map.get(nums[i])+1); }else{ map.put(nums[i], 1); } } // System.out.println(map); PriorityQueue<Element> pb = new PriorityQueue<Element>(); for(Map.Entry<Integer,Integer> entry : map.entrySet()){ pb.add(new Element(entry.getKey(), entry.getValue())); } for (int i = 0; i < k; i++) { list.add(pb.poll().getNum()); } return list; } } class Element implements Comparable<Element>{ private Integer num; private Integer count; public Element(int num, int count) { super(); this.num = num; this.count = count; } public Integer getNum() { return num; } public void setNum(Integer num) { this.num = num; } public Integer getCount() { return count; } public void setCount(Integer count) { this.count = count; } @Override public int compareTo(Element o) { return o.count.compareTo(this.count); } }