力扣692.前K个高频单词

692.前K个高频单词

思路1：

1. 创建HashMap，并把单词以及其频率存进HashMap中
2. 将keySet()存进List链表中，并用Collections的sort进行逻辑的排序
3. 再创建满足题意的新链表，返回新链表即可

代码实现

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
		HashMap<String, Integer> map = new HashMap<String, Integer>();
		for (int i = 0; i < words.length; i++) {
			int frequency = map.getOrDefault(words[i], 0) + 1;
			map.put(words[i], frequency);
		}
		List<String> list = new LinkedList<String>(map.keySet());
		Collections.sort(list, new Comparator<String>() {

			@Override
			public int compare(String o1, String o2) {
				if (map.get(o1).equals(map.get(o2))) {
					return o1.compareTo(o2);
				} else {
					return map.get(o2) - map.get(o1);
				}
			}
		});
		List<String> list2 = new LinkedList<String>();
		for (int i = 0; i < k; i++) {
			list2.add(list.get(i));
		}
		return list2;
    }
}

思路2:

1. 创建HashMap并存进去
2. 使用优先队列，改成最大堆以及题目相应的逻辑
3. 取k次堆顶即可

代码实现

class Solution {
    public List<String> topKFrequent(String[] words, int k) {
        HashMap<String, Integer> map = new HashMap<>();
		for (String str : words) {
			map.put(str, map.getOrDefault(str, 0) + 1);
		}
		
		PriorityQueue<String> queue = new PriorityQueue<>(new Comparator<String>() {

			@Override
			public int compare(String o1, String o2) {
				if (map.get(o1).equals(map.get(o2))) {
					return o1.compareTo(o2);
				} else {
					return map.get(o2) - map.get(o1);
				}
			}
		});
        for (String word : map.keySet()) {
			queue.add(word);
		}
		List<String> list = new LinkedList<String>();
		for (int i = 0; i < k; i++) {
			list.add(queue.poll());
		}
		return list;
    }
}