![https://leetcode.com/problems/remove-outermost-parentheses/]
A valid parentheses string is either empty (""), “(” + A + “)”, or A + B, where A and B are valid parentheses strings, and + represents string concatenation. For example, “”, “()”, “(())()”, and “(()(()))” are all valid parentheses strings.
A valid parentheses string S is primitive if it is nonempty, and there does not exist a way to split it into S = A+B, with A and B nonempty valid parentheses strings.
Given a valid parentheses string S, consider its primitive decomposition: S = P_1 + P_2 + … + P_k, where P_i are primitive valid parentheses strings.
Return S after removing the outermost parentheses of every primitive string in the primitive decomposition of S.
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
| Example 1:
Input: "(()())(())"
Output: "()()()"
Explanation:
The input string is "(()())(())", with primitive decomposition "(()())" + "(())".
After removing outer parentheses of each part, this is "()()" + "()" = "()()()".
Example 2:
Input: "(()())(())(()(()))"
Output: "()()()()(())"
Explanation:
The input string is "(()())(())(()(()))", with primitive decomposition "(()())" + "(())" + "(()(()))".
After removing outer parentheses of each part, this is "()()" + "()" + "()(())" = "()()()()(())".
Example 3:
Input: "()()"
Output: ""
Explanation:
The input string is "()()", with primitive decomposition "()" + "()".
After removing outer parentheses of each part, this is "" + "" = "".
|
Note:
- S.length <= 10000
- S[i] is “(” or “)”
- S is a valid parentheses string
1
2
3
4
5
6
7
8
9
10
11
12
13
14
15
16
17
18
19
20
21
22
23
24
25
26
27
28
29
30
31
32
33
34
35
36
37
38
39
40
41
42
43
44
45
46
47
48
49
50
51
52
53
54
55
56
| class Solution {
/*
Explanation:
opened count the number of opened parenthesis.
Add every char to the result,
unless the first left parenthesis,
and the last right parenthesis.
"( ()() ) (())"
1
*/
public String removeOuterParentheses(String S) {
StringBuilder sb = new StringBuilder();
int opened = 0;
for (char c : S.toCharArray()) {
if (c == '(' && opened++ > 0) sb.append(c);
if (c == ')' && opened-- > 1) sb.append(c);
}
return sb.toString();
}
public String removeOuterParentheses2(String S) {
char[] symbols = S.toCharArray();
int count = 0;
List<Integer> stack = new ArrayList<>();
for (int i = 0; i < symbols.length; i++) {
char c = symbols[i];
if (c == '(') {
stack.add(i);
count++;
} else {
Integer removed = null;
if (!stack.isEmpty() && symbols[stack.get(stack.size() - 1)] == '(') {
removed = stack.get(stack.size() - 1);
stack.remove(stack.size() - 1);
} else {
stack.add(i);
}
count--;
if (count == 0) {
symbols[i] = '_';
symbols[removed] = '_';
}
}
}
StringBuilder sb = new StringBuilder();
for (char c : symbols) {
if (c != '_') {
sb.append(c);
}
}
return sb.toString();
}
}
|