7373
7474### 方法一:模拟
7575
76- 用双端队列或者栈,模拟反转的过程 。
76+ 我们可以直接用栈来模拟反转的过程 。
7777
78- 时间复杂度 $O(n^2),ドル其中 $n$ 为字符串 $s$ 的长度。
78+ 时间复杂度 $O(n^2),ドル空间复杂度 $O(n),ドル 其中 $n$ 为字符串 $s$ 的长度。
7979
8080<!-- tabs:start -->
8181
@@ -86,46 +86,37 @@ class Solution:
8686 def reverseParentheses (self s : str ) -> str :
8787 stk = []
8888 for c in s:
89- if c == ' ) '
89+ if c == " ) "
9090 t = []
91- while stk[- 1 ] != ' ( '
91+ while stk[- 1 ] != " ( "
9292 t.append(stk.pop())
9393 stk.pop()
9494 stk.extend(t)
9595 else :
9696 stk.append(c)
97- return ' '
97+ return " "
9898```
9999
100100#### Java
101101
102102``` java
103103class Solution {
104104 public String reverseParentheses (String s ) {
105- int n = s. length();
106- int [] d = new int [n];
107- Deque<Integer > stk = new ArrayDeque<> ();
108- for (int i = 0 ; i < n; ++ i) {
109- if (s. charAt(i) == ' ('
110- stk. push(i);
111- } else if (s. charAt(i) == ' )'
112- int j = stk. pop();
113- d[i] = j;
114- d[j] = i;
115- }
116- }
117- StringBuilder ans = new StringBuilder ();
118- int i = 0 , x = 1 ;
119- while (i < n) {
120- if (s. charAt(i) == ' (' || s. charAt(i) == ' )'
121- i = d[i];
122- x = - x;
105+ StringBuilder stk = new StringBuilder ();
106+ for (char c : s. toCharArray()) {
107+ if (c == ' )'
108+ StringBuilder t = new StringBuilder ();
109+ while (stk. charAt(stk. length() - 1 ) != ' ('
110+ t. append(stk. charAt(stk. length() - 1 ));
111+ stk. deleteCharAt(stk. length() - 1 );
112+ }
113+ stk. deleteCharAt(stk. length() - 1 );
114+ stk. append(t);
123115 } else {
124- ans . append(s . charAt(i) );
116+ stk . append(c );
125117 }
126- i += x;
127118 }
128- return ans . toString();
119+ return stk . toString();
129120 }
130121}
131122```
@@ -177,6 +168,27 @@ func reverseParentheses(s string) string {
177168}
178169```
179170
171+ #### TypeScript
172+ 173+ ``` ts
174+ function reverseParentheses(s : string ): string {
175+ const : string [] = [];
176+ for (const of s ) {
177+ if (c === ' )'
178+ const : string [] = [];
179+ while (stk .at (- 1 )! !== ' ('
180+ t .push (stk .pop ()! );
181+ }
182+ stk .pop ();
183+ stk .push (... t );
184+ } else {
185+ stk .push (c );
186+ }
187+ }
188+ return stk .join (' '
189+ }
190+ ```
191+ 180192#### JavaScript
181193
182194``` js
@@ -185,32 +197,20 @@ func reverseParentheses(s string) string {
185197 * @return {string}
186198 */
187199var reverseParentheses = function (s ) {
188- const n = s .length ;
189- const d = new Array (n).fill (0 );
190200 const stk = [];
191- for (let i = 0 ; i < n; ++ i) {
192- if (s[i] == ' ('
193- stk .push (i);
194- } else if (s[i] == ' )'
195- const j = stk .pop ();
196- d[i] = j;
197- d[j] = i;
198- }
199- }
200- let i = 0 ;
201- let x = 1 ;
202- const ans = [];
203- while (i < n) {
204- const c = s .charAt (i);
205- if (c == ' (' || c == ' )'
206- i = d[i];
207- x = - x;
201+ for (const c of s) {
202+ if (c === ' )'
203+ const t = [];
204+ while (stk .at (- 1 ) !== ' ('
205+ t .push (stk .pop ());
206+ }
207+ stk .pop ();
208+ stk .push (... t);
208209 } else {
209- ans .push (c);
210+ stk .push (c);
210211 }
211- i += x;
212212 }
213- return ans .join (' '
213+ return stk .join (' '
214214};
215215```
216216
@@ -228,7 +228,7 @@ var reverseParentheses = function (s) {
228228
229229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。
230230
231- 时间复杂度 $O(n),ドル其中 $n$ 为字符串 $s$ 的长度。
231+ 时间复杂度 $O(n),ドル空间复杂度 $O(n),ドル 其中 $n$ 为字符串 $s$ 的长度。
232232
233233<!-- tabs: start -->
234234
@@ -241,21 +241,54 @@ class Solution:
241241 d = [0 ] * n
242242 stk = []
243243 for i, c in enumerate (s):
244- if c == ' ( '
244+ if c == " ( "
245245 stk.append(i)
246- elif c == ' ) '
246+ elif c == " ) "
247247 j = stk.pop()
248248 d[i], d[j] = j, i
249249 i, x = 0 , 1
250250 ans = []
251251 while i < n:
252- if s[i] in ' () '
252+ if s[i] in " () "
253253 i = d[i]
254254 x = - x
255255 else :
256256 ans.append(s[i])
257257 i += x
258- return ' '
258+ return " "
259+ ```
260+ 261+ #### Java
262+ 263+ ``` java
264+ class Solution {
265+ public String reverseParentheses (String s ) {
266+ int n = s. length();
267+ int [] d = new int [n];
268+ Deque<Integer > stk = new ArrayDeque<> ();
269+ for (int i = 0 ; i < n; ++ i) {
270+ if (s. charAt(i) == ' ('
271+ stk. push(i);
272+ } else if (s. charAt(i) == ' )'
273+ int j = stk. pop();
274+ d[i] = j;
275+ d[j] = i;
276+ }
277+ }
278+ StringBuilder ans = new StringBuilder ();
279+ int i = 0 , x = 1 ;
280+ while (i < n) {
281+ if (s. charAt(i) == ' (' || s. charAt(i) == ' )'
282+ i = d[i];
283+ x = - x;
284+ } else {
285+ ans. append(s. charAt(i));
286+ }
287+ i += x;
288+ }
289+ return ans. toString();
290+ }
291+ }
259292```
260293
261294#### C++
@@ -324,6 +357,76 @@ func reverseParentheses(s string) string {
324357}
325358```
326359
360+ #### TypeScript
361+ 362+ ``` ts
363+ function reverseParentheses(s : string ): string {
364+ const = s .length ;
365+ const : number [] = Array (n ).fill (0 );
366+ const : number [] = [];
367+ for (let i = 0 ; i < n ; ++ i ) {
368+ if (s [i ] === ' ('
369+ stk .push (i );
370+ } else if (s [i ] === ' )'
371+ const = stk .pop ()! ;
372+ d [i ] = j ;
373+ d [j ] = i ;
374+ }
375+ }
376+ let i = 0 ;
377+ let x = 1 ;
378+ const : string [] = [];
379+ while (i < n ) {
380+ const = s .charAt (i );
381+ if (' ()' includes (c )) {
382+ i = d [i ];
383+ x = - x ;
384+ } else {
385+ ans .push (c );
386+ }
387+ i += x ;
388+ }
389+ return ans .join (' '
390+ }
391+ ```
392+ 393+ #### JavaScript
394+ 395+ ``` js
396+ /**
397+ * @param {string} s
398+ * @return {string}
399+ */
400+ var reverseParentheses = function (s ) {
401+ const n = s .length ;
402+ const d = Array (n).fill (0 );
403+ const stk = [];
404+ for (let i = 0 ; i < n; ++ i) {
405+ if (s[i] === ' ('
406+ stk .push (i);
407+ } else if (s[i] === ' )'
408+ const j = stk .pop ();
409+ d[i] = j;
410+ d[j] = i;
411+ }
412+ }
413+ let i = 0 ;
414+ let x = 1 ;
415+ const ans = [];
416+ while (i < n) {
417+ const c = s .charAt (i);
418+ if (' ()' includes (c)) {
419+ i = d[i];
420+ x = - x;
421+ } else {
422+ ans .push (c);
423+ }
424+ i += x;
425+ }
426+ return ans .join (' '
427+ };
428+ ```
429+ 327430<!-- tabs: end -->
328431
329432<!-- solution: end -->
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