LeetCode 921. Minimum Add to Make Parentheses Valid

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Given a string S of '(' and ')' parentheses, we add the minimum number of parentheses ( '(' or ')', and in any positions ) so that the resulting parentheses string is valid.

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

• It is the empty string, 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.

Given a parentheses string, return the minimum number of parentheses we must add to make the resulting string valid.

Example 1:

Input: "())"
Output: 1

Example 2:

Input: "((("
Output: 3

Example 3:

Input: "()"
Output: 0

Example 4:

Input: "()))(("
Output: 4

Note:

1. S.length <= 1000
2. S only consists of '(' and ')' characters.

此题我的理解为找出其不能成对的括号即可，将其加入一个栈中，成对则弹出，剩余则为不能成对的。Accepted代码如下：

class Solution {
    public int minAddToMakeValid(String S) {
        int l = S.length();
        if (l == 0) {
            return 0;
        }
        if (l == 1) {
            return 1;
        }
        Stack<Character> stack = new Stack();
        for (int i = 0; i < l; i++) {
            char a = S.charAt(i);
            if (a == '(') {
                stack.push(a);
            } else if (stack.empty()){
                stack.push(a);
            } else if ('(' == stack.peek()) {
                stack.pop();
            } else {
                stack.push(a);
            }
        }
        return stack.size();
    }
}