优先队列解决的问题

src/main/java/com/ximo/datastructuresinaction/leetcode/solution347/Solution.java

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+package com.ximo.datastructuresinaction.leetcode.solution347;
+
+import com.ximo.datastructuresinaction.queue.PriorityQueue;
+
+import java.util.*;
+
+import static java.util.stream.Collectors.counting;
+import static java.util.stream.Collectors.groupingBy;
+
+/**
+ * @author xikl
+ * @date 2019/1/5
+ */
+public class Solution {
+
+    private class Freq implements Comparable<Freq> {
+        /** 元素的值 */
+        private Integer element;
+
+        /** 元素出现个数 */
+        private Long freq;
+
+        public Freq(Integer element, Long freq) {
+            this.element = element;
+            this.freq = freq;
+        }
+
+        public Freq() {
+        }
+
+        /** 因为自己的优先队列是最大堆 这里的判断大小要反过来 */
+        @Override
+        public int compareTo(Freq another) {
+//            if (this.freq < another.freq) {
+//                return 1;
+//            } else if (this.freq > another.freq) {
+//                return -1;
+//            } else {
+//                return 0;
+//            }
+            // 上面的代码可以改写为这样
+            Long compareResult = another.freq - this.freq;
+            return compareResult.intValue();
+        }
+    }
+
+    public List<Integer> topKFrequent(int[] nums, int k) {
+        return topKFrequentByMyQueue(nums, k);
+    }
+
+    public List<Integer> topKFrequentByMyQueue(int[] nums, int k) {
+        // 将nums按照出现频率变为一个map
+        Map<Integer, Long> numFrequency =
+                Arrays.stream(nums).boxed().collect(groupingBy(num -> num, counting()));
+
+        PriorityQueue<Freq> pq = new PriorityQueue<>();
+
+        numFrequency.forEach((key, value) -> {
+            if (pq.getSize() < k) {
+                pq.enqueue(new Freq(key, value));
+            } else if (value > pq.getFront().freq){
+                pq.dequeue();
+                pq.enqueue(new Freq(key, value));
+            }
+        });
+
+        List<Integer> result = new LinkedList<>();
+        while (!pq.isEmpty()) {
+            result.add(pq.dequeue().element);
+        }
+
+        return result;
+    }
+
+    /** 用java的自带的优先队列实现 */
+    public List<Integer> topKFrequentByJavaQueue(int[] nums, int k) {
+        // 将nums按照出现频率变为一个map
+        Map<Integer, Long> numFrequency =
+                Arrays.stream(nums).boxed().collect(groupingBy(num -> num, counting()));
+
+        // java中采用的是小顶堆的实现
+        java.util.Queue<Integer> pq = new java.util.PriorityQueue<>((prev , next) -> {
+            Long compareResult = numFrequency.get(prev) - numFrequency.get(next);
+            return compareResult.intValue();
+        });
+
+        numFrequency.forEach((key, value) -> {
+            if (pq.size() < k) {
+                pq.add(key);
+            } else if (value > numFrequency.get(pq.peek())) {
+                pq.remove();
+                pq.add(key);
+            }
+        });
+        return new ArrayList<>(pq);
+    }
+
+
+
+}

src/test/java/com/ximo/datastructuresinaction/map/MapTest.java

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+package com.ximo.datastructuresinaction.map;
+
+import org.junit.Test;
+
+import java.util.Arrays;
+import java.util.Map;
+import java.util.function.Function;
+
+import static java.util.stream.Collectors.counting;
+import static java.util.stream.Collectors.groupingBy;
+
+/**
+ * @author xikl
+ * @date 2019/1/5
+ */
+public class MapTest {
+
+
+    @Test
+    public void testGroupingBy() {
+        Integer[] nums = {1, 1, 1, 2, 2, 3};
+        Map<Integer, Long> collect = Arrays.stream(nums).collect(groupingBy(Function.identity(), counting()));
+        System.out.println(collect);
+    }
+}