[Swift]LeetCode1190. 反转每对括号间的子串 | Reverse Substrings Between Each Pair of Parentheses

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Given a string s that consists of lower case English letters and brackets. 

Reverse the strings in each pair of matching parentheses, starting from the innermost one.

Your result should not contain any bracket.

Example 1:

Input: s = "(abcd)"
Output: "dcba"

Example 2:

Input: s = "(u(love)i)"
Output: "iloveu"

Example 3:

Input: s = "(ed(et(oc))el)"
Output: "leetcode"

Example 4:

Input: s = "a(bcdefghijkl(mno)p)q"
Output: "apmnolkjihgfedcbq"

Constraints:

• 0 <= s.length <= 2000
• s only contains lower case English characters and parentheses.
• It's guaranteed that all parentheses are balanced.

---

给出一个字符串 s（仅含有小写英文字母和括号）。

请你按照从括号内到外的顺序，逐层反转每对匹配括号中的字符串，并返回最终的结果。

注意，您的结果中 不应 包含任何括号。

示例 1：

输入：s = "(abcd)"
输出："dcba"

示例 2：

输入：s = "(u(love)i)"
输出："iloveu"

示例 3：

输入：s = "(ed(et(oc))el)"
输出："leetcode"

示例 4：

输入：s = "a(bcdefghijkl(mno)p)q"
输出："apmnolkjihgfedcbq"

提示：

• 0 <= s.length <= 2000
• s 中只有小写英文字母和括号
• 我们确保所有括号都是成对出现的

---

Runtime: 8 ms

Memory Usage: 20.9 MB

class Solution {
    func reverseParentheses(_ s: String) -> String {
        var s1:[String] = [String]()
        var s2:[Character] = [Character]()
        var sb:String = String()
        var s = Array(s)
        for i in 0..<s.count
        {
            let c:Character = s[i]
            if c == "("
            {
                s2.append(c)
                s1.append(sb)
                sb = String()
            }
            else if c == ")"
            {
                s2.removeLast()
                sb = String(sb.reversed())
                if !s1.isEmpty
                {
                    sb.insert(contentsOf: s1.removeLast(), at:sb.startIndex)
                }
            }
            else
            {
                sb.append(c)
            }
        }
        return sb
    }
}