top k frequent elements solution

Author: Yn3-3xh

top-k-frequent-elements/Yn3-3xh.java

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+/*
+[문제풀이]
+time: O(N log N), space: O(N + k)
+- 주어진 Nums의 중복된 수의 갯수를 Map에 담고,
+- 역정렬하면,
+- k만큼 뽑기
+class Solution1 {
+    public int[] topKFrequent(int[] nums, int k) {
+        Map<Integer, Integer> counting = new HashMap<>();
+        for (int num: nums) {
+            int numCount = counting.getOrDefault(num, 0) + 1;
+            counting.put(num, numCount);
+        }
+
+        List<Map.Entry<Integer, Integer>> sortedCounting = new ArrayList<>(counting.entrySet());
+        sortedCounting.sort(Map.Entry.comparingByValue(Comparator.reverseOrder()));
+
+        int[] result = new int[k];
+        for (int i = 0; i < k; i++) {
+            result[i] = sortedCounting.get(i).getKey();
+        }
+
+        return result;
+    }
+}
+
+time: O(N log k), space: O(N + k)
+- PriorityQueue 사용
+- value로 역정렬 설정
+- key로 queue에 add
+
+[회고]
+첫번째 풀이는 역정렬 하는 부분이 미숙했고,
+두번째 풀이는 PriorityQueue를 사용하는데 처음 써본다.. (익히자!)
+*/
+class Solution {
+    public int[] topKFrequent(int[] nums, int k) {
+        Map<Integer, Integer> counting = new HashMap<>();
+        for (int num: nums) {
+            int numCount = counting.getOrDefault(num, 0) + 1;
+            counting.put(num, numCount);
+        }
+
+        Queue<Integer> queue = new PriorityQueue<>((count1, count2) ->
+                counting.get(count2) - counting.get(count1));
+        queue.addAll(counting.keySet());
+
+        int[] result = new int[k];
+        for (int i = 0; i < k; i++) {
+            result[i] = queue.poll();
+        }
+        return result;
+    }
+}