77 lines
2.2 KiB
Java
77 lines
2.2 KiB
Java
package heap;
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import java.util.HashMap;
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import java.util.Map;
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import java.util.PriorityQueue;
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/**
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* 题目： 347. 前 K 个高频元素 (topKFrequent)
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* 描述：给你一个整数数组 nums 和一个整数 k ，请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。
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* 示例 1：
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输入: nums = [1,1,1,2,2,3], k = 2
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输出: [1,2]
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* 链接：https://leetcode.cn/problems/top-k-frequent-elements/
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*/
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// 二刷不会
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public class TopKFrequent {
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public int[] topKFrequent(int[] nums, int k) {
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HashMap<Integer,Integer> map=new HashMap<>();
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int cnt=0;
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PriorityQueue<int[]> minHeap=new PriorityQueue<>( //小根堆
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(a,b) -> a[1]-b[1]
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);
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int[] temp;
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int[] res=new int[k];
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for (int i = 0; i < nums.length; i++) {
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int cur=nums[i];
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map.put(cur, map.getOrDefault(cur,0)+1);
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}
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for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
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if(cnt<k){
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minHeap.add(new int[]{entry.getKey(),entry.getValue()});
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cnt++;
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}else {
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temp=minHeap.peek();
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if(entry.getValue()>temp[1]) {
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minHeap.poll();
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minHeap.add(new int[]{entry.getKey(),entry.getValue()});
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}
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}
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}
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cnt=0;
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for (int[] ints : minHeap) {
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res[cnt++]=ints[0];
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}
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return res;
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}
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public int[] topKFrequent2(int[] nums, int k) {
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HashMap<Integer,Integer>map=new HashMap<>();
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PriorityQueue<int[]> maxHeap = new PriorityQueue<>( //大根堆
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(a, b) -> b[1] - a[1]);
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int[] res=new int[k];
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int i=0;
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for (int num : nums) {
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if(!map.containsKey(num))
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map.put(num,1);
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else
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map.put(num,map.get(num)+1);
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}
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for (Map.Entry<Integer, Integer> entry : map.entrySet()) {
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maxHeap.add(new int[]{entry.getKey(),entry.getValue()});
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}
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while (!maxHeap.isEmpty()) {
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if(i<k)
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res[i++] = maxHeap.poll()[0];
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else break;
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}
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return res;
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}
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}
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