题目描述:

解:
先用Map统计词频,然后放入集合类按需排序取前k个即可。
class Solution {public List<String> topKFrequent(String[] words, int k) {HashMap<String,Integer> map = new HashMap<String, Integer>();for(String word : words) {map.put(word,map.getOrDefault(word,0)+1);}List<String> ret = new ArrayList<String>();for(Map.Entry<String, Integer> entry : map.entrySet()) {ret.add(entry.getKey());}Collections.sort(ret,(word1,word2) -> {return map.get(word1) == map.get(word2) ? word1.compareTo(word2) : map.get(word2) - map.get(word1);});return ret.subList(0,k);}}
进阶:
小根堆优化
public class Solution {public List<String> topKFrequent(String[] words, int k) {// 1.先用哈希表统计单词出现的频率Map<String, Integer> count = new HashMap();for (String word : words) {count.put(word, count.getOrDefault(word, 0) + 1);}// 2.构建小根堆 这里需要自己构建比较规则 此处为 lambda 写法 Java 的优先队列默认实现就是小根堆PriorityQueue<String> minHeap = new PriorityQueue<>((s1, s2) -> {if (count.get(s1).equals(count.get(s2))) {return s2.compareTo(s1);} else {return count.get(s1) - count.get(s2);}});// 3.依次向堆加入元素。for (String s : count.keySet()) {minHeap.offer(s);// 当堆中元素个数大于 k 个的时候,需要弹出堆顶最小的元素。if (minHeap.size() > k) {minHeap.poll();}}// 4.依次弹出堆中的 K 个元素,放入结果集合中。List<String> res = new ArrayList<String>(k);while (minHeap.size() > 0) {res.add(minHeap.poll());}// 5.注意最后需要反转元素的顺序。Collections.reverse(res);return res;}}作者:qi-ye-jun链接:https://leetcode-cn.com/problems/top-k-frequent-words/solution/xiao-gen-dui-huo-zhe-hashbiao-pai-xu-by-9uj06/来源:力扣(LeetCode)著作权归作者所有。商业转载请联系作者获得授权,非商业转载请注明出处。
