347. Top K Frequent Elements（找出数组中出现次数最多的前k个元素）

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].

Note: 

• You may assume k is always valid, 1 ≤ k ≤ number of unique elements.
• Your algorithm's time complexity must be better than O(n log n), where n is the array's size.

题目大意：在一个整型数组中，按出现次数从多到少找出前k个元素，将其对应的值存入一个list中返回。

解题思路：

解法一：根据元素出现的次数进行堆排序。

代码如下：（34ms，beats 53.21%）

public List topKFrequent(int[] nums, int k) {
		List res = new LinkedList();
		Map map = new HashMap();
		for (int num : nums) {
			if (map.get(num) == null)
				map.put(num, 1);
			else
				map.put(num, (int) (map.get(num)) + 1);
		}
		int len = map.size();
		Entry[] entrys = new Entry[len + 1];
		Iterator> it = map.entrySet().iterator();
		for (int i = 1; i <= len; i++)
			entrys[i] = (Entry) it.next();

		for (int i = len / 2; i >= 1; i--) {
			adjust(entrys, len, i);
		}
		for (int i = 1; i <= k; i++) {
			res.add((Integer) entrys[1].getKey());
			entrys[1] = entrys[len];
			len--;
			adjust(entrys, len, 1);
		}
		return res;
	}

	public void adjust(Entry[] entrys, int len, int i) {
		int k;
		entrys[0] = entrys[i];
		for (k = i * 2; k <= len; k *= 2) {
			if (k + 1 <= len && entrys[k].getValue() < entrys[k + 1].getValue())
				k = k + 1;
			if (entrys[0].getValue() < entrys[k].getValue()) {
				entrys[i] = entrys[k];
				i = k;
			} else
				break;
		}
		entrys[i] = entrys[0];
	}

解法二：桶排序。

代码如下：（30ms，beats 72.58%）

public List topKFrequent2(int[] nums, int k) {
		List res = new LinkedList();
		Map map = new HashMap();
		for (int num : nums) {
			if (map.get(num) == null)
				map.put(num, 1);
			else
				map.put(num, (int) (map.get(num)) + 1);
		}
		int len = nums.length;
		ArrayList[] times = new ArrayList[len + 1];
		Set set = map.keySet();
		Iterator it = set.iterator();
		while (it.hasNext()) {
			int num = it.next();
			int time = map.get(num);
			if (times[time] == null)
				times[time] = new ArrayList<>();
			times[map.get(num)].add(num);
		}
		for (int i = len; i >= 0; i--)
			if (times[i] != null) {
				it = times[i].iterator();
				while (k > 0 && it.hasNext()) {
					res.add(it.next());
					k--;
				}
			}
		return res;
	}

解法三：利用Collections.sort函数

代码如下：（36ms，beats 43.93%）

public  List topKFrequent3(int[] nums, int k) {
			List res = new LinkedList();
			Map map = new HashMap();
			for (int num : nums) {
				if (map.get(num) == null)
					map.put(num, 1);
				else
					map.put(num, (int) (map.get(num)) + 1);
			}
			List> list = new ArrayList>(map.entrySet());
			Collections.sort(list, new Comparator>() {

				@Override
				public int compare(Entry o1, Entry o2) {
					return o2.getValue()-o1.getValue();
				}
				
			});
			for(int i=0;i