feat: add js solution to lc problem: No.1190

Author: rain84

solution/1100-1199/1190.Reverse Substrings Between Each Pair of Parentheses/README.md

@@ -357,6 +357,44 @@ func reverseParentheses(s string) string {
 }
 ```
 
+#### JavaScript
+
+```js
+function reverseParentheses(s: string): string {
+    const res: string[] = [];
+    const n = s.length;
+    const pairs = Array(n).fill(-1);
+    const stack: number[] = [];
+
+    for (let i = 0; i < n; i++) {
+        if (s[i] === '(') stack.push(i);
+        else if (s[i] === ')') {
+            const j = stack.pop()!;
+            pairs[i] = j;
+            pairs[j] = i;
+        }
+    }
+
+    for (let i = 0, forward = true; i < n; ) {
+        const ch = s[i];
+
+        switch (s[i]) {
+            case '(':
+            case ')':
+                i = forward ? pairs[i] - 1 : pairs[i] + 1;
+                forward = !forward;
+                break;
+
+            default:
+                res.push(ch);
+                i = forward ? i + 1 : i - 1;
+        }
+    }
+
+    return res.join('');
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1100-1199/1190.Reverse Substrings Between Each Pair of Parentheses/README_EN.md

@@ -350,6 +350,44 @@ func reverseParentheses(s string) string {
 }
 ```
 
+#### JavaScript
+
+```js
+function reverseParentheses(s: string): string {
+    const res: string[] = [];
+    const n = s.length;
+    const pairs = Array(n).fill(-1);
+    const stack: number[] = [];
+
+    for (let i = 0; i < n; i++) {
+        if (s[i] === '(') stack.push(i);
+        else if (s[i] === ')') {
+            const j = stack.pop()!;
+            pairs[i] = j;
+            pairs[j] = i;
+        }
+    }
+
+    for (let i = 0, forward = true; i < n; ) {
+        const ch = s[i];
+
+        switch (s[i]) {
+            case '(':
+            case ')':
+                i = forward ? pairs[i] - 1 : pairs[i] + 1;
+                forward = !forward;
+                break;
+
+            default:
+                res.push(ch);
+                i = forward ? i + 1 : i - 1;
+        }
+    }
+
+    return res.join('');
+}
+```
+
 <!-- tabs:end -->
 
 <!-- solution:end -->

solution/1100-1199/1190.Reverse Substrings Between Each Pair of Parentheses/Solution2.js

@@ -0,0 +1,33 @@
+function reverseParentheses(s) {
+    const res = [];
+    const n = s.length;
+    const pairs = Array(n).fill(-1);
+    const stack = [];
+
+    for (let i = 0; i < n; i++) {
+        if (s[i] === '(') stack.push(i);
+        else if (s[i] === ')') {
+            const j = stack.pop();
+            pairs[i] = j;
+            pairs[j] = i;
+        }
+    }
+
+    for (let i = 0, forward = true; i < n; ) {
+        const ch = s[i];
+
+        switch (s[i]) {
+            case '(':
+            case ')':
+                i = forward ? pairs[i] - 1 : pairs[i] + 1;
+                forward = !forward;
+                break;
+
+            default:
+                res.push(ch);
+                i = forward ? i + 1 : i - 1;
+        }
+    }
+
+    return res.join('');
+}