diff --git a/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/README.md b/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/README.md
index 41990ec30..2648c06e2 100755
--- a/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/README.md
+++ b/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/README.md
@@ -1,28 +1,38 @@
 # [2182.Construct String With Repeat Limit][title]
 
-> [!WARNING|style:flat]
-> This question is temporarily unanswered if you have good ideas. Welcome to [Create Pull Request PR](https://github.com/kylesliu/awesome-golang-algorithm)
-
 ## Description
+You are given a string `s` and an integer `repeatLimit`. Construct a new string `repeatLimitedString` using the characters of `s` such that no letter appears **more than** `repeatLimit` times **in a row**. You do **not** have to use all characters from `s`.
+
+Return the **lexicographically largest** `repeatLimitedString` possible.
+
+A string `a` is **lexicographically larger** than a string `b` if in the first position where `a` and `b` differ, string `a` has a letter that appears later in the alphabet than the corresponding letter in `b`. If the first `min(a.length, b.length)` characters do not differ, then the longer string is the lexicographically larger one.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "cczazcc", repeatLimit = 3
+Output: "zzcccac"
+Explanation: We use all of the characters from s to construct the repeatLimitedString "zzcccac".
+The letter 'a' appears at most 1 time in a row.
+The letter 'c' appears at most 3 times in a row.
+The letter 'z' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "zzcccac".
+Note that the string "zzcccca" is lexicographically larger but the letter 'c' appears more than 3 times in a row, so it is not a valid repeatLimitedString.
 ```
 
-## 题意
-> ...
+**Example 2:**
 
-## 题解
-
-### 思路1
-> ...
-Construct String With Repeat Limit
-```go
 ```
-
+Input: s = "aababab", repeatLimit = 2
+Output: "bbabaa"
+Explanation: We use only some of the characters from s to construct the repeatLimitedString "bbabaa". 
+The letter 'a' appears at most 2 times in a row.
+The letter 'b' appears at most 2 times in a row.
+Hence, no letter appears more than repeatLimit times in a row and the string is a valid repeatLimitedString.
+The string is the lexicographically largest repeatLimitedString possible so we return "bbabaa".
+Note that the string "bbabaaa" is lexicographically larger but the letter 'a' appears more than 2 times in a row, so it is not a valid repeatLimitedString.
+```
 
 ## 结语
 
diff --git a/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution.go b/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution.go
index d115ccf5e..ff4064b50 100644
--- a/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution.go
+++ b/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution.go
@@ -1,5 +1,66 @@
 package Solution
 
-func Solution(x bool) bool {
+import (
+	"bytes"
+	"container/heap"
+)
+
+type tmp2182 struct {
+	b byte
+	c int
+}
+type heap2182 []tmp2182
+
+func (h *heap2182) Len() int {
+	return len(*h)
+}
+
+func (h *heap2182) Less(i, j int) bool {
+	return (*h)[i].b > (*h)[j].b
+}
+
+func (h *heap2182) Swap(i, j int) {
+	(*h)[i], (*h)[j] = (*h)[j], (*h)[i]
+}
+
+func (h *heap2182) Push(x any) {
+	*h = append(*h, x.(tmp2182))
+}
+
+func (h *heap2182) Pop() any {
+	old := *h
+	l := len(old)
+	x := old[l-1]
+	*h = old[:l-1]
 	return x
 }
+
+func Solution(s string, repeatLimit int) string {
+	c := make(map[byte]int)
+	for _, b := range []byte(s) {
+		c[b]++
+	}
+	list := &heap2182{}
+	for k, v := range c {
+		heap.Push(list, tmp2182{k, v})
+	}
+	buf := bytes.NewBuffer([]byte{})
+
+	for list.Len() > 0 {
+		top := heap.Pop(list).(tmp2182)
+		for range min(top.c, repeatLimit) {
+			buf.WriteByte(top.b)
+		}
+		top.c -= repeatLimit
+		if top.c > 0 && list.Len() > 0 {
+			nextPickOne := heap.Pop(list).(tmp2182)
+			buf.WriteByte(nextPickOne.b)
+			nextPickOne.c--
+			if nextPickOne.c > 0 {
+				heap.Push(list, nextPickOne)
+			}
+			heap.Push(list, top)
+		}
+	}
+	return buf.String()
+}
diff --git a/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution_test.go b/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution_test.go
index 14ff50eb4..061653c6e 100644
--- a/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution_test.go
+++ b/leetcode/2101-2200/2182.Construct-String-With-Repeat-Limit/Solution_test.go
@@ -10,30 +10,30 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		s      string
+		limit  int
+		expect string
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "cczazcc", 3, "zzcccac"},
+		{"TestCase2", "aababab", 2, "bbabaa"},
 	}
 
 	//	开始测试
 	for i, c := range cases {
 		t.Run(c.name+" "+strconv.Itoa(i), func(t *testing.T) {
-			got := Solution(c.inputs)
+			got := Solution(c.s, c.limit)
 			if !reflect.DeepEqual(got, c.expect) {
-				t.Fatalf("expected: %v, but got: %v, with inputs: %v",
-					c.expect, got, c.inputs)
+				t.Fatalf("expected: %v, but got: %v, with inputs: %v %v",
+					c.expect, got, c.s, c.limit)
 			}
 		})
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }