diff --git a/Algorithms/347. Top K Frequent Elements/347. Top K Frequent Elements.go b/Algorithms/347. Top K Frequent Elements/347. Top K Frequent Elements.go
new file mode 100644
index 00000000..04d47fd4
--- /dev/null
+++ b/Algorithms/347. Top K Frequent Elements/347. Top K Frequent Elements.go	
@@ -0,0 +1,53 @@
+package leetcode
+
+import "container/heap"
+
+func topKFrequent(nums []int, k int) []int {
+	m := make(map[int]int)
+	for _, n := range nums {
+		m[n]++
+	}
+	q := PriorityQueue{}
+	for key, count := range m {
+		heap.Push(&q, &Item{key: key, count: count})
+	}
+	var result []int
+	for len(result) < k {
+		item := heap.Pop(&q).(*Item)
+		result = append(result, item.key)
+	}
+	return result
+}
+
+type Item struct {
+	key   int
+	count int
+}
+
+// A PriorityQueue implements heap.Interface and holds Items.
+type PriorityQueue []*Item
+
+func (pq PriorityQueue) Len() int {
+	return len(pq)
+}
+
+func (pq PriorityQueue) Less(i, j int) bool {
+	// 注意：因为golang中的heap是按最小堆组织的，所以count越大，Less()越小，越靠近堆顶.
+	return pq[i].count > pq[j].count
+}
+
+func (pq PriorityQueue) Swap(i, j int) {
+	pq[i], pq[j] = pq[j], pq[i]
+}
+
+func (pq *PriorityQueue) Push(x interface{}) {
+	item := x.(*Item)
+	*pq = append(*pq, item)
+}
+
+func (pq *PriorityQueue) Pop() interface{} {
+	n := len(*pq)
+	item := (*pq)[n-1]
+	*pq = (*pq)[:n-1]
+	return item
+}
diff --git a/Algorithms/347. Top K Frequent Elements/347. Top K Frequent Elements_test.go b/Algorithms/347. Top K Frequent Elements/347. Top K Frequent Elements_test.go
new file mode 100644
index 00000000..8d28f9ad
--- /dev/null
+++ b/Algorithms/347. Top K Frequent Elements/347. Top K Frequent Elements_test.go	
@@ -0,0 +1,48 @@
+package leetcode
+
+import (
+	"fmt"
+	"testing"
+)
+
+type question347 struct {
+	para347
+	ans347
+}
+
+// para 是参数
+// one 代表第一个参数
+type para347 struct {
+	one []int
+	two int
+}
+
+// ans 是答案
+// one 代表第一个答案
+type ans347 struct {
+	one []int
+}
+
+func Test_Problem347(t *testing.T) {
+
+	qs := []question347{
+
+		question347{
+			para347{[]int{1, 1, 1, 2, 2, 3}, 2},
+			ans347{[]int{1, 2}},
+		},
+
+		question347{
+			para347{[]int{1}, 1},
+			ans347{[]int{1}},
+		},
+	}
+
+	fmt.Printf("------------------------Leetcode Problem 347------------------------\n")
+
+	for _, q := range qs {
+		_, p := q.ans347, q.para347
+		fmt.Printf("【input】:%v       【output】:%v\n", p, topKFrequent(p.one, p.two))
+	}
+	fmt.Printf("\n\n\n")
+}
diff --git a/Algorithms/347. Top K Frequent Elements/README.md b/Algorithms/347. Top K Frequent Elements/README.md
new file mode 100644
index 00000000..5db21c2d
--- /dev/null
+++ b/Algorithms/347. Top K Frequent Elements/README.md	
@@ -0,0 +1,33 @@
+# [347. Top K Frequent Elements](https://leetcode.com/problems/top-k-frequent-elements/)
+
+## 题目
+
+Given a non-empty array of integers, return the k most frequent elements.
+
+Example 1:
+
+```c
+Input: nums = [1,1,1,2,2,3], k = 2
+Output: [1,2]
+```
+
+Example 2:
+
+```c
+Input: nums = [1], k = 1
+Output: [1]
+```
+
+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
+
+给一个非空的数组，输出前 K 个频率最高的元素。
+
+这一题是考察优先队列的题目。把数组构造成一个优先队列，输出前 K 个即可。
+
+