1249. Minimum Remove to Make Valid Parentheses-Toy模板网

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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:文章来源地址https://www.toymoban.com/news/detail-744824.html

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;
        
    }
}

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