原题链接在这里：https://leetcode.com/problems/minimum-remove-to-make-valid-parentheses/

题目：

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)"

Constraints:

• 1 <= s.length <= 10^5
• s[i] is one of  '(' , ')' and lowercase English letters.

题解：

From left to right, accoulate left open parenthese. 

When encountering ")" and there is no more left "(", then this should be ignored.

Do the same thing from right left.

Time Complexity: O(n). n = s.length().

Space: O(n).

AC Java:

 1 class Solution {
 2     public String minRemoveToMakeValid(String s) {
 3         if(s == null || s.length() == 0){
 4             return s;
 5         }
 6         
 7         StringBuilder sb = new StringBuilder();
 8         int left = 0;
 9         for(char c : s.toCharArray()){
10             if(c == '('){
11                 left++;
12             }else if(c == ')'){
13                 if(left == 0){
14                     continue;
15                 }
16                 
17                 left--;
18             }
19             
20             sb.append(c);
21         }
22         
23         StringBuilder res = new StringBuilder();
24         for(int i = sb.length()-1; i>=0; i--){
25             char c = sb.charAt(i);
26             if(c == '(' && left-- > 0){
27                 continue;
28             }
29             
30             res.append(c);
31         }
32         
33         return res.reverse().toString();
34     }
35 }