From 555dfe5590534b78944127e379602741b5408d68 Mon Sep 17 00:00:00 2001
From: huangge1199 <huangge1199@hotmail.com>
Date: Fri, 3 Sep 2021 13:43:47 +0800
Subject: [PATCH] =?UTF-8?q?347:=E5=89=8D=20K=20=E4=B8=AA=E9=AB=98=E9=A2=91?=
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---
 .../editor/cn/TopKFrequentElements.java       | 77 +++++++++++++++++++
 .../cn/doc/content/TopKFrequentElements.md    | 31 ++++++++
 2 files changed, 108 insertions(+)
 create mode 100644 src/main/java/leetcode/editor/cn/TopKFrequentElements.java
 create mode 100644 src/main/java/leetcode/editor/cn/doc/content/TopKFrequentElements.md

diff --git a/src/main/java/leetcode/editor/cn/TopKFrequentElements.java b/src/main/java/leetcode/editor/cn/TopKFrequentElements.java
new file mode 100644
index 0000000..28041d4
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/TopKFrequentElements.java
@@ -0,0 +1,77 @@
+//给你一个整数数组 nums 和一个整数 k ，请你返回其中出现频率前 k 高的元素。你可以按 任意顺序 返回答案。 
+//
+// 
+//
+// 示例 1: 
+//
+// 
+//输入: nums = [1,1,1,2,2,3], k = 2
+//输出: [1,2]
+// 
+//
+// 示例 2: 
+//
+// 
+//输入: nums = [1], k = 1
+//输出: [1] 
+//
+// 
+//
+// 提示： 
+//
+// 
+// 1 <= nums.length <= 10⁵ 
+// k 的取值范围是 [1, 数组中不相同的元素的个数] 
+// 题目数据保证答案唯一，换句话说，数组中前 k 个高频元素的集合是唯一的 
+// 
+//
+// 
+//
+// 进阶：你所设计算法的时间复杂度 必须 优于 O(n log n) ，其中 n 是数组大小。 
+// Related Topics 数组 哈希表 分治 桶排序 计数 快速选择 排序 堆（优先队列） 👍 853 👎 0
+
+package leetcode.editor.cn;
+
+import java.util.*;
+
+//347:前 K 个高频元素
+class TopKFrequentElements {
+    public static void main(String[] args) {
+        //测试代码
+        Solution solution = new TopKFrequentElements().new Solution();
+        solution.topKFrequent(new int[]{1, 1, 1, 2, 2, 3}, 2);
+    }
+
+    //力扣代码
+    //leetcode submit region begin(Prohibit modification and deletion)
+    class Solution {
+        public int[] topKFrequent(int[] nums, int k) {
+            Map<Integer, Integer> counts = new HashMap<>();
+            for (int num : nums) {
+                counts.put(num, counts.getOrDefault(num, 0) + 1);
+            }
+            Map<Integer, List<Integer>> map = new HashMap<>();
+            for (int key : counts.keySet()) {
+                List<Integer> temp = map.getOrDefault(counts.get(key), new ArrayList<>());
+                temp.add(key);
+                map.put(counts.get(key), new ArrayList<>(temp));
+            }
+            List<Integer> list = new ArrayList<>(map.keySet());
+            int[] arr = new int[k];
+            int index = 0;
+            for (int i = list.size() - 1; i >= 0; i--) {
+                List<Integer> temp = map.get(list.get(i));
+                for (int j = index; j < index + temp.size(); j++) {
+                    arr[j] = temp.get(j - index);
+                }
+                index += temp.size();
+                if (index == k) {
+                    break;
+                }
+            }
+            return arr;
+        }
+    }
+//leetcode submit region end(Prohibit modification and deletion)
+
+}
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diff --git a/src/main/java/leetcode/editor/cn/doc/content/TopKFrequentElements.md b/src/main/java/leetcode/editor/cn/doc/content/TopKFrequentElements.md
new file mode 100644
index 0000000..b139850
--- /dev/null
+++ b/src/main/java/leetcode/editor/cn/doc/content/TopKFrequentElements.md
@@ -0,0 +1,31 @@
+<p>给你一个整数数组 <code>nums</code> 和一个整数 <code>k</code> ，请你返回其中出现频率前 <code>k</code> 高的元素。你可以按 <strong>任意顺序</strong> 返回答案。</p>
+
+<p> </p>
+
+<p><strong>示例 1:</strong></p>
+
+<pre>
+<strong>输入: </strong>nums = [1,1,1,2,2,3], k = 2
+<strong>输出: </strong>[1,2]
+</pre>
+
+<p><strong>示例 2:</strong></p>
+
+<pre>
+<strong>输入: </strong>nums = [1], k = 1
+<strong>输出: </strong>[1]</pre>
+
+<p> </p>
+
+<p><strong>提示：</strong></p>
+
+<ul>
+	<li><code>1 <= nums.length <= 10<sup>5</sup></code></li>
+	<li><code>k</code> 的取值范围是 <code>[1, 数组中不相同的元素的个数]</code></li>
+	<li>题目数据保证答案唯一，换句话说，数组中前 <code>k</code> 个高频元素的集合是唯一的</li>
+</ul>
+
+<p> </p>
+
+<p><strong>进阶：</strong>你所设计算法的时间复杂度 <strong>必须</strong> 优于 <code>O(n log n)</code> ，其中 <code>n</code><em> </em>是数组大小。</p>
+<div><div>Related Topics</div><div><li>数组</li><li>哈希表</li><li>分治</li><li>桶排序</li><li>计数</li><li>快速选择</li><li>排序</li><li>堆（优先队列）</li></div></div><br><div><li>👍 853</li><li>👎 0</li></div>
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