1249. Minimum Remove to Make Valid Parentheses-CSDN博客

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.

Constraints:

• 1 <= s.length <= 105
• s[i] is either'(' , ')', or lowercase English letter.

class Solution {
    public String minRemoveToMakeValid(String s) {
        if(s == null || s.length() == 0){
            return "";
        }
        //做括号的题一定要用栈切记。这题用的是贪心算法找到左括号进栈找到右括号出栈进栈出栈的都是char array的index。
        
        char[] ans = s.toCharArray();
                
        Stack<Integer> stack = new Stack<Integer>();
        
        for(int i=0; i<ans.length; i++){
            if(ans[i] == '('){
                stack.push(i);
            }
            if(ans[i] == ')'){
                if(stack.isEmpty()){ //如果这个时候栈为空更改右括号为空。
                    ans[i] = ' ';
                }else{
                    stack.pop();
                }
            }
        }
        while(!stack.isEmpty()){ //如果最后还剩左括号把左括号都置为空
            int index = stack.pop();
            ans[index] = ' ';
        }
        
        String result = new String(ans);
        
        result = result.replace(" ", "");
        
        return result;
        
    }
}