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Jul 12, 2024
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feat: add ts/js solutions to lc problem: No.1190 #3260

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feat: add ts solution to lc problem: No.1190
rain84 Jul 11, 2024
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feat: add js solution to lc problem: No.1190
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feat: add ts solution to lc problem: No.1190
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feat: update ts/js solutions to lc problem: No.1190
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209 changes: 156 additions & 53 deletions ...on/1100-1199/1190.Reverse Substrings Between Each Pair of Parentheses/README.md
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Original file line number Diff line number Diff line change
Expand Up @@ -73,9 +73,9 @@ tags:

### 方法一：模拟

用双端队列或者栈，模拟反转的过程。
我们可以直接用栈来模拟反转的过程。

时间复杂度 $O(n^2)$，其中 $n$ 为字符串 $s$ 的长度。
时间复杂度 $O(n^2)$，空间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。

<!-- tabs:start -->

Expand All @@ -86,46 +86,37 @@ class Solution:
def reverseParentheses(self, s: str) -> str:
stk = []
for c in s:
if c == ')':
if c == ")":
t = []
while stk[-1] != '(':
while stk[-1] != "(":
t.append(stk.pop())
stk.pop()
stk.extend(t)
else:
stk.append(c)
return ''.join(stk)
return "".join(stk)
```

#### Java

```java
class Solution {
public String reverseParentheses(String s) {
int n = s.length();
int[] d = new int[n];
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == '(') {
stk.push(i);
} else if (s.charAt(i) == ')') {
int j = stk.pop();
d[i] = j;
d[j] = i;
}
}
StringBuilder ans = new StringBuilder();
int i = 0, x = 1;
while (i < n) {
if (s.charAt(i) == '(' || s.charAt(i) == ')') {
i = d[i];
x = -x;
StringBuilder stk = new StringBuilder();
for (char c : s.toCharArray()) {
if (c == ')') {
StringBuilder t = new StringBuilder();
while (stk.charAt(stk.length() - 1) != '(') {
t.append(stk.charAt(stk.length() - 1));
stk.deleteCharAt(stk.length() - 1);
}
stk.deleteCharAt(stk.length() - 1);
stk.append(t);
} else {
ans.append(s.charAt(i));
stk.append(c);
}
i += x;
}
return ans.toString();
return stk.toString();
}
}
```
Expand Down Expand Up @@ -177,6 +168,27 @@ func reverseParentheses(s string) string {
}
```

#### TypeScript

```ts
function reverseParentheses(s: string): string {
const stk: string[] = [];
for (const c of s) {
if (c === ')') {
const t: string[] = [];
while (stk.at(-1)! !== '(') {
t.push(stk.pop()!);
}
stk.pop();
stk.push(...t);
} else {
stk.push(c);
}
}
return stk.join('');
}
```

#### JavaScript

```js
Expand All @@ -185,32 +197,20 @@ func reverseParentheses(s string) string {
* @return {string}
*/
var reverseParentheses = function (s) {
const n = s.length;
const d = new Array(n).fill(0);
const stk = [];
for (let i = 0; i < n; ++i) {
if (s[i] == '(') {
stk.push(i);
} else if (s[i] == ')') {
const j = stk.pop();
d[i] = j;
d[j] = i;
}
}
let i = 0;
let x = 1;
const ans = [];
while (i < n) {
const c = s.charAt(i);
if (c == '(' || c == ')') {
i = d[i];
x = -x;
for (const c of s) {
if (c === ')') {
const t = [];
while (stk.at(-1) !== '(') {
t.push(stk.pop());
}
stk.pop();
stk.push(...t);
} else {
ans.push(c);
stk.push(c);
}
i += x;
}
return ans.join('');
return stk.join('');
};
```

Expand All @@ -228,7 +228,7 @@ var reverseParentheses = function (s) {

然后，我们从左到右遍历字符串，遇到 `(` 或者 `)` 时，根据 $d$ 数组跳到对应的位置，然后反转方向，继续遍历，直到遍历完整个字符串。

时间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。
时间复杂度 $O(n)$，空间复杂度 $O(n)$，其中 $n$ 为字符串 $s$ 的长度。

<!-- tabs:start -->

Expand All @@ -241,21 +241,54 @@ class Solution:
d = [0] * n
stk = []
for i, c in enumerate(s):
if c == '(':
if c == "(":
stk.append(i)
elif c == ')':
elif c == ")":
j = stk.pop()
d[i], d[j] = j, i
i, x = 0, 1
ans = []
while i < n:
if s[i] in '()':
if s[i] in "()":
i = d[i]
x = -x
else:
ans.append(s[i])
i += x
return ''.join(ans)
return "".join(ans)
```

#### Java

```java
class Solution {
public String reverseParentheses(String s) {
int n = s.length();
int[] d = new int[n];
Deque<Integer> stk = new ArrayDeque<>();
for (int i = 0; i < n; ++i) {
if (s.charAt(i) == '(') {
stk.push(i);
} else if (s.charAt(i) == ')') {
int j = stk.pop();
d[i] = j;
d[j] = i;
}
}
StringBuilder ans = new StringBuilder();
int i = 0, x = 1;
while (i < n) {
if (s.charAt(i) == '(' || s.charAt(i) == ')') {
i = d[i];
x = -x;
} else {
ans.append(s.charAt(i));
}
i += x;
}
return ans.toString();
}
}
```

#### C++
Expand Down Expand Up @@ -324,6 +357,76 @@ func reverseParentheses(s string) string {
}
```

#### TypeScript

```ts
function reverseParentheses(s: string): string {
const n = s.length;
const d: number[] = Array(n).fill(0);
const stk: number[] = [];
for (let i = 0; i < n; ++i) {
if (s[i] === '(') {
stk.push(i);
} else if (s[i] === ')') {
const j = stk.pop()!;
d[i] = j;
d[j] = i;
}
}
let i = 0;
let x = 1;
const ans: string[] = [];
while (i < n) {
const c = s.charAt(i);
if ('()'.includes(c)) {
i = d[i];
x = -x;
} else {
ans.push(c);
}
i += x;
}
return ans.join('');
}
```

#### JavaScript

```js
/**
* @param {string} s
* @return {string}
*/
var reverseParentheses = function (s) {
const n = s.length;
const d = Array(n).fill(0);
const stk = [];
for (let i = 0; i < n; ++i) {
if (s[i] === '(') {
stk.push(i);
} else if (s[i] === ')') {
const j = stk.pop();
d[i] = j;
d[j] = i;
}
}
let i = 0;
let x = 1;
const ans = [];
while (i < n) {
const c = s.charAt(i);
if ('()'.includes(c)) {
i = d[i];
x = -x;
} else {
ans.push(c);
}
i += x;
}
return ans.join('');
};
```

<!-- tabs:end -->

<!-- solution:end -->
Expand Down
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