LeetCode 1021. Remove Outermost Parentheses
原题链接在这里:https://leetcode.com/problems/remove-outermost-parentheses/
题目:
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
.
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
题解:
Count open parentheses, when it is (, count++. If it is outermost (, count == 1 now and skip.
When it is ), count--. If it is outermost ), count == 0 now and skip.
Otherwise append it to StringBuilder.
Time Complexity: O(n). n = S.length().
Space: O(n).
AC Java:
1 class Solution { 2 public String removeOuterParentheses(String S) { 3 int count = 0; 4 StringBuilder sb = new StringBuilder(); 5 for(int i = 0; i<S.length(); i++){ 6 char c = S.charAt(i); 7 if(c == '('){ 8 count++; 9 if(count == 1){ 10 continue; 11 } 12 }else{ 13 count--; 14 if(count == 0){ 15 continue; 16 } 17 } 18 19 sb.append(c); 20 } 21 22 return sb.toString(); 23 } 24 }