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  • [LeetCode] 1021. 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.

    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:

    1. S.length <= 10000
    2. S[i] is "(" or ")"
    3. S is a valid parentheses string

    删除最外层的括号。

    有效括号字符串为空 ("")、"(" + A + ")" 或 A + B,其中 A 和 B 都是有效的括号字符串,+ 代表字符串的连接。例如,"","()","(())()" 和 "(()(()))" 都是有效的括号字符串。

    如果有效字符串 S 非空,且不存在将其拆分为 S = A+B 的方法,我们称其为原语(primitive),其中 A 和 B 都是非空有效括号字符串。

    给出一个非空有效字符串 S,考虑将其进行原语化分解,使得:S = P_1 + P_2 + ... + P_k,其中 P_i 是有效括号字符串原语。

    对 S 进行原语化分解,删除分解中每个原语字符串的最外层括号,返回 S 。

    来源:力扣(LeetCode)
    链接:https://leetcode-cn.com/problems/remove-outermost-parentheses
    著作权归领扣网络所有。商业转载请联系官方授权,非商业转载请注明出处。

    这道题不难,但是题目描述解释了半天,只有最后一句话是有用的,就是移除input字符串中每个括号对的最外层的一对括号。

    思路是用类似stack的思想去track到底有几层,只有当层数 >= 1的时候才需要保存遍历到的括号。

    时间O(n)

    空间O(1)

    Java实现

     1 class Solution {
     2     public String removeOuterParentheses(String S) {
     3         StringBuilder s = new StringBuilder();
     4         int opened = 0;
     5         for (char c : S.toCharArray()) {
     6             if (c == '(' && opened++ > 0) {
     7                 s.append(c);
     8             }
     9             if (c == ')' && opened-- > 1) {
    10                 s.append(c);
    11             }
    12         }
    13         return s.toString();
    14     }
    15 }

    LeetCode 题目总结

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  • 原文地址:https://www.cnblogs.com/cnoodle/p/14182318.html
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