1249. Minimum Remove to Make Valid Parentheses

Given a string s of ‘(‘ , ‘)‘ and lowercase English characters.

Your task is to remove the minimum number of parentheses ( ‘(‘ or ‘)‘, in any positions ) so that the resulting parentheses string is valid and return any valid string.

Formally, a parentheses string is valid if and only if:

• It is the empty string, contains only lowercase characters, or
• It can be written as AB (A concatenated with B), where A and B are valid strings, or
• It can be written as (A), where A is a valid string.

Example 1:

Input: s = "lee(t(c)o)de)"
Output: "lee(t(c)o)de"
Explanation: "lee(t(co)de)" , "lee(t(c)ode)" would also be accepted.

Example 2:

Input: s = "a)b(c)d"
Output: "ab(c)d"

Example 3:

Input: s = "))(("
Output: ""
Explanation: An empty string is also valid.

Example 4:

Input: s = "(a(b(c)d)"
Output: "a(b(c)d)"

分析：用一个stack来记录需要删除的括号，最后把它们删除了就可以了。

 1 class Solution {
 2     public String minRemoveToMakeValid(String s) {
 3       StringBuilder sb = new StringBuilder(s);
 4       Stack<Integer> st = new Stack();
 5       for (int i = 0; i < sb.length(); ++i) {
 6         if (sb.charAt(i) == ‘(‘) {
 7             st.add(i + 1);
 8         } else if (sb.charAt(i) == ‘)‘) {
 9           if (!st.empty() && st.peek() > 0) {
10               st.pop();
11           } else {
12               st.add(-(i + 1));
13           }
14         }
15       }
16       while (!st.empty()) {
17         sb.deleteCharAt(Math.abs(st.pop()) - 1);
18       }
19       return sb.toString();
20     }
21 }

原文地址：https://www.cnblogs.com/beiyeqingteng/p/12268458.html