diff --git a/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md b/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md
index b58cd9741..1ae7dc123 100644
--- a/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md
+++ b/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/README.md
@@ -1,28 +1,34 @@
 # [1190.Reverse Substrings Between Each Pair of Parentheses][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` that consists of lower case English letters and brackets.
+
+Reverse the strings in each pair of matching parentheses, starting from the innermost one.
+
+Your result should **not** contain any brackets.
 
 **Example 1:**
 
 ```
-Input: a = "11", b = "1"
-Output: "100"
+Input: s = "(abcd)"
+Output: "dcba"
 ```
 
-## 题意
-> ...
-
-## 题解
+**Example 2:**
 
-### 思路1
-> ...
-Reverse Substrings Between Each Pair of Parentheses
-```go
 ```
+Input: s = "(u(love)i)"
+Output: "iloveu"
+Explanation: The substring "love" is reversed first, then the whole string is reversed.
+```
+
+**Example 3:**
 
+```
+Input: s = "(ed(et(oc))el)"
+Output: "leetcode"
+Explanation: First, we reverse the substring "oc", then "etco", and finally, the whole string.
+```
 
 ## 结语
 
diff --git a/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution.go b/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution.go
index d115ccf5e..33c409735 100644
--- a/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution.go
+++ b/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution.go
@@ -1,5 +1,40 @@
 package Solution
 
-func Solution(x bool) bool {
-	return x
+import (
+	"bytes"
+)
+
+func Solution(s string) string {
+	stack := make([][]byte, 0)
+	buf := bytes.NewBuffer([]byte{})
+	for idx := 0; idx < len(s); idx++ {
+		if s[idx] == '(' {
+			dst := make([]byte, len(buf.Bytes()))
+			copy(dst, buf.Bytes())
+
+			stack = append(stack, dst)
+			buf.Reset()
+			continue
+		}
+		if s[idx] == ')' {
+			tmp := buf.Bytes()
+			tmpl := len(tmp)
+			bs := make([]byte, tmpl)
+			buf.Reset()
+			for i := tmpl - 1; i >= 0; i-- {
+				bs[tmpl-i-1] = tmp[i]
+			}
+
+			l := len(stack) - 1
+			top := stack[l]
+			stack = stack[:l]
+
+			top = append(top, bs...)
+			buf.Write(top)
+			continue
+		}
+		buf.WriteByte(s[idx])
+	}
+
+	return buf.String()
 }
diff --git a/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution_test.go b/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution_test.go
index 14ff50eb4..96dcd82b7 100644
--- a/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution_test.go
+++ b/leetcode/1101-1200/1190.Reverse-Substrings-Between-Each-Pair-of-Parentheses/Solution_test.go
@@ -10,12 +10,12 @@ func TestSolution(t *testing.T) {
 	//	测试用例
 	cases := []struct {
 		name   string
-		inputs bool
-		expect bool
+		inputs string
+		expect string
 	}{
-		{"TestCase", true, true},
-		{"TestCase", true, true},
-		{"TestCase", false, false},
+		{"TestCase1", "(u(love)i)", "iloveu"},
+		{"TestCase2", "(abcd)", "dcba"},
+		{"TestCase3", "(ed(et(oc))el)", "leetcode"},
 	}
 
 	//	开始测试
@@ -30,10 +30,10 @@ func TestSolution(t *testing.T) {
 	}
 }
 
-//	压力测试
+// 压力测试
 func BenchmarkSolution(b *testing.B) {
 }
 
-//	使用案列
+// 使用案列
 func ExampleSolution() {
 }