Leetcode692 Top K Frequent Words

题目描述

example:

Input: ["i", "love", "leetcode", "i", "love", "coding"], k = 2
Output: ["i", "love"]
Explanation: "i" and "love" are the two most frequent words.
    Note that "i" comes before "love" due to a lower alphabetical order.

Input: ["the", "day", "is", "sunny", "the", "the", "the", "sunny", "is", "is"], k = 4
Output: ["the", "is", "sunny", "day"]
Explanation: "the", "is", "sunny" and "day" are the four most frequent words,
    with the number of occurrence being 4, 3, 2 and 1 respectively.

解决方案：

Step1: 统计词频：

	Map<String, Integer> map = new HashMap();
    for (String word: words) {
        map.put(word, map.getOrDefault(word, 0) + 1);
    }
或：
    HashMap<String, Integer> map = new HashMap<>();
    for(String word: words){
        if(map.containsKey(word)){
            map.put(word,map.get(word)+1);
        }else
            map.put(word,1); 
    }

Step2: 比较：

List<String> result = new ArrayList<>();
for(String key : map.keySet()){
   result.add(key);
}
Collections.sort(result,(word1,word2)->map.get(word1).equals(map.get(word2))?
                 word1.compareTo(word2):map.get(word2)-map.get(word1));

Step3: 输出：

return result.subList(0,k);

最优解：Heap

Step1: 统计词频：

Map<String, Integer> map = new HashMap();
for (String word: words) {
    map.put(word, map.getOrDefault(word, 0) + 1);
}

Step2: PriorityQueue:

PriorityQueue<String> heap = new PriorityQueue<String>(
                (w1, w2) -> count.get(w1).equals(count.get(w2)) ?
                w2.compareTo(w1) : count.get(w1) - count.get(w2) );

Step3: 从heap提取前k个值，然后倒序输出

for (String word: count.keySet()) {
           heap.offer(word);
           if (heap.size() > k) heap.poll();
       }

       List<String> ans = new ArrayList();
       while (!heap.isEmpty()) ans.add(heap.poll());
       Collections.reverse(ans);
       return ans;